Chapter 6 Application of Derivatives

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NCERT Solutions Class 12 Maths Chapter 6 Application of Derivatives

Class 12 Maths Chapter 6: Application of Derivatives focuses on using derivatives and explains how derivatives help in finding the rate of change, increasing and decreasing functions, tangents and normals, maxima and minima, and practical problems based on optimisation. These concepts show how calculus can be applied beyond formulas and is a very important topic from an exam point of view.

ALLEN provides Class 12 Maths Chapter 6 NCERT Solutions prepared by subject experts according to the latest textbook and aligned with the current syllabus prescribed by CBSE. The solutions are explained step by step in simple language so that Class 12 students can easily understand applications and methods. Practising NCERT Solutions on Application of Derivatives regularly helps improve analytical thinking, accuracy, and confidence for board and competitive exams.

1.0Key Concepts of NCERT Solutions Class 12 Maths Chapter 6 Applications of Derivatives

Class 12 Maths Chapter 6, Applications of Derivatives, explains how derivatives can be used to study the behaviour of functions and solve different mathematical problems. The main topics covered in NCERT Solutions for Class 12 Maths Chapter 6 include:

Rate of Change: Understand how derivatives represent the rate at which one quantity changes with respect to another.
Increasing and Decreasing Functions: Learn how derivatives can be used to identify the intervals where a function is increasing or decreasing.
Tangents and Normals: Learn how to find the equations of tangents and normals to a curve at a given point.
Maxima and Minima: Understand how derivatives are used to find the maximum and minimum values of a function.
Applications in Real-Life Problems: Apply derivatives to solve optimisation and other practical problems involving changing quantities.

2.0NCERT Class 12 Maths Chapter 6 Applications of Derivatives : Detailed Solutions

EXERCISE - 6.1

Find the rate of change of the area of a circle with respect to its radius r when (a) $r = 3 cm$ (b) $r = 4 cm$ Sol. The area of a circle (A) with radius ( r ) is given by, $A = π r^{2}$ Now, the rate of change of the area with respect to its radius is given by,

$\frac{dA}{dr} = \frac{d}{dr} (π r^{2}) = 2 π r$

(a) When $r = 3 cm, \frac{dA}{dr} = 2 π (3) = 6 π$ Hence, the area of the circle is changing at the rate of $6 π cm^{2} / cm$ when its radius is 3 cm. (b) When $r = 4 cm, \frac{dA}{dr} = 2 π (4) = 8 π$ Hence, the area of the circle is changing at the rate of $8 π cm^{2} / cm$ when its radius is 4 cm.

The volume of a cube is increasing at the rate of $8 cm^{3} / s$ . How fast is the surface area increasing when the length of an edge is 12 cm? Sol. Let x be the length of side, V be the volume and S be the surface area of the cube. Then, $V = x^{3}$ and $S = 6 x^{2}$ It is given that $\frac{dV}{dt} = 8 cm^{3} / s$ Then, by using the chain rule, we have :

$∴ \Rightarrow 8 = \frac{dV}{dt} = \frac{d}{dt} (x^{3}) = 3 x^{2} \cdot \frac{dx}{dt} \frac{dx}{dt} = \frac{8}{3 x ^{2}}$

Now, $\frac{dS}{dt} = \frac{d}{dt} (6 x^{2}) = \frac{d}{dx} (6 x^{2}) . \frac{dx}{dt}$ [By chain rule]

$= 12 x \cdot \frac{dx}{dt} = 12 x \cdot (\frac{8}{3 x ^{2}}) = \frac{32}{x}$

[From (1)] Thus, when $x = 12 cm$ ,

$\frac{dS}{dt} = \frac{32}{12} cm^{2} / s = \frac{8}{3} cm^{2} / s$

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of $\frac{8}{3} cm^{2} / s$ .

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm. Sol. The area of a circle (A) with radius ( r ) is given by,

$A = π r^{2}$

Now, the rate of change of area (A) with respect to time (t) is given by,

$\frac{dA}{dt} = \frac{d}{dt} (π r^{2}) = 2 π r \frac{dr}{dt} [By chain rule]$

It is given that, $\frac{dr}{dt} = 3 cm / s$ $∴ \frac{dA}{dt} = 2 π r (3) = 6 π r$

Thus, when $r = 10 cm$ ,

$\frac{dA}{dt} = 6 π (10) = 60 π cm^{2} / s$

Hence, the rate at which the area of the circle is increasing, when the radius is 10 cm, is $60 π cm^{2} / s$ .

An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long? Sol. Let x be the length of a side and V be the volume of the cube. Then, $V = x^{3}$ .

$∴ \frac{dV}{dt} = 3 x^{2} \cdot \frac{dx}{dt} [By chain rule]$

It is given that, $\frac{dx}{dt} = 3 cm / s$

$∴ \frac{dV}{dt} = 3 x^{2} (3) = 9 x^{2}$

Therefore, when $x = 10 cm$ ,

$\frac{dV}{dt} = 9 (10)^{2} = 900 cm^{3} / s$

Hence, the volume of the cube is increasing at the rate of $900 cm^{3} / s$ when the edge is 10 cm long.

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing? Sol. The area of a circle (A) with radius ( r ) is given by $A = π r^{2}$ . Therefore, the rate of change of area (A) with respect to time (t) is given by,

$\frac{dA}{dt} = \frac{d}{dt} (π r^{2}) = \frac{d}{dr} (π r^{2}) \frac{dr}{dt} = 2 π r \frac{dr}{dt}$

[By chain rule] It is given that $\frac{dr}{dt} = 5 cm / s$ Thus, when $r = 8 cm, \frac{dA}{dt} = 2 π (8) (5) = 80 π$ Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of $80 π cm^{2} / s$ .

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? Sol. The circumference of a circle (C) with radius $(r)$ is given by $C = 2 π r$ . Therefore, the rate of change of circumference (C) with respect to time (t) is given by,

$\frac{dC}{dt} = \frac{dC}{dr} \cdot \frac{dr}{dt} [By chain rule] = \frac{d}{dr} (2 π r) \frac{dr}{dt} = 2 π \cdot \frac{dr}{dt}$

It is given that $\frac{dr}{dt} = 0.7 cm / s$ Hence, the rate of increase of the circumference is $2 π (0.7) = 1.4 π cm / s$ .

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When $x = 8 cm$ and $y = 6 cm$ , find the rates of change of (a) the perimeter, and (b) the area of the rectangle. Sol. Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

$\frac{dx}{dt} = - 5 cm / min and \frac{dy}{dt} = 4 cm / min, x = 8 cm and y = 6 cm$

(a) The perimeter (P) of a rectangle is given by, $P = 2 (x + y)$ $∴ \frac{dP}{dt} = 2 (\frac{dx}{dt} + \frac{dy}{dt}) = 2 (- 5 + 4) = - 2 cm / min$

(Differentiate w.r.t. t) Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by, $A = x \cdot y$ $∴ \frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt} = - 5 y + 4 x$ (Differentiate w.r.t. t) When $x = 8 cm$ and $y = 6 cm$ , $\frac{dA}{dt} = (- 5 \times 6 + 4 \times 8) cm^{2} / min = 2 cm^{2} / min$ Hence, the area of the rectangle is increasing at the rate of $2 cm^{2} / min$ .

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. Sol. The volume of a sphere (V) with radius (r) is given by, $V = \frac{4}{3} π r^{3}$ ∴ Rate of change of volume (V) with respect to time (t) is given by,

$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} [Chain rule] = \frac{d}{dr} (\frac{4}{3} π r^{3}) \cdot \frac{dr}{dt} = 4 π r^{2} \cdot \frac{dr}{dt}$

It is given that $\frac{dV}{dt} = 900 cm^{3} / s$

$∴ 900 = 4 π r^{2} \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{900}{4 π r ^{2}} = \frac{225}{π r ^{2}}$ Therefore, when radius $= 15 cm$ ,

$\frac{dr}{dt} = \frac{225}{π ( 15 ) ^{2}} = \frac{1}{π}$

Hence, the rate at which the radius of the balloon increases, when the radius is 15 cm, is $\frac{1}{π} cm / s$ .

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm. Sol. The volume of a sphere (V) with radius (r) is given by $V = \frac{4}{3} π r^{3}$ . Rate of change of volume (V) with respect to its radius (r) is given by,

$\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} π r^{3}) = \frac{4}{3} π (3 r^{2}) = 4 π r^{2}$

Therefore when radius $= 10 cm$ ,

$\frac{dV}{dr} = 4 π (10)^{2} = 400 π cm^{3} / cm .$

Hence, the volume of the balloon is increasing at the rate of $400 π cm^{3} / cm$ .

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall? Sol. Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall. Then, by Pythagoras theorem, we have:

$x^{2} + y^{2} = 25 [Length of the ladder = 5 m]$

$\Rightarrow y = 25 - x^{2}$

Then, the rate of change of height (y) with respect to time ( t ) is given by

$\frac{dy}{dt} = \frac{- x}{25 - x ^{2}} \cdot \frac{dx}{dt}$

It is given that $\frac{dx}{dt} = 0.02 m / s$ . $∴ \frac{dy}{dt} = \frac{- x}{25 - x ^{2}} \times \frac{2}{100} m / s$ Now when $x = 4 m$ , we have : $\frac{dy}{dt} = \frac{- 4 \times 2}{100 \times 3} m / s = - \frac{8}{300} m / s = - \frac{8}{3} cm / s$ Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{8}{3} cm / s$ .

A particle moves along the curve $6 y = x^{3} + 2$ . Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. Sol. The equation of the curve is given as:

$6 y = x^{3} + 2$

The rate of change of the position of the particle with respect to time (t) is given by

$6 \frac{dy}{dt} = 3 x^{2} \frac{dx}{dt} + 0$

$\Rightarrow 2 \frac{dy}{dt} = x^{2} \frac{dx}{dt}$

When the y -coordinate of the particle changes 8 times as fast as the x -coordinate

i.e., $(\frac{dy}{dt} = 8 \frac{dx}{dt})$ , we have:

$2 (8 \frac{dx}{dt}) = x^{2} \frac{dx}{dt}$

$\Rightarrow 16 \frac{dx}{dt} = x^{2} \frac{dx}{dt}$ $\Rightarrow (x^{2} - 16) \frac{dx}{dt} = 0$

$\Rightarrow x^{2} = 16 \Rightarrow x = \pm 4$

When $x = 4, y = \frac{4 ^{3} + 2}{6} = \frac{66}{6} = 11$ When $x = - 4, y = \frac{( - 4 ) ^{3} + 2}{6} = - \frac{62}{6} = - \frac{31}{3}$ Hence, the points on the curve are $(4, 11)$ and

$(- 4, \frac{- 31}{3})$

The radius of an air bubble is increasing at the rate of $\frac{1}{2} cm / s$ . At what rate is the volume of the bubble increasing when the radius is 1 cm? Sol. The air bubble is in the shape of a sphere. Now, the volume of an air bubble (V) with radius (r) is given by, $V = \frac{4}{3} π r^{3}$ The rate of change of volume (V) with respect to time (t) is given by,

$\frac{dV}{dt} = \frac{4}{3} π \frac{d}{dr} (r^{3}) \cdot \frac{dr}{dt} [By chain rule] = \frac{4}{3} π (3 r^{2}) \frac{dr}{dt} = 4 π r^{2} \frac{dr}{dt}$

It is given that $\frac{dr}{dt} = \frac{1}{2} cm / s$ Therefore, when $r = 1 cm$ ,

$\frac{dV}{dt} = 4 π (1)^{2} (\frac{1}{2}) = 2 π cm^{3} / s$

Hence, the rate at which the volume of the bubble increases is $2 π cm^{3} / s$ .

A balloon, which always remains spherical, has a variable diameter $\frac{3}{2} (2 x + 1)$ . Find the rate of change of its volume with respect to x. Sol. The volume of a sphere (V) with radius (r) is given by, $V = \frac{4}{3} π r^{3}$ It is given that:

$Diameter = \frac{3}{2} (2 x + 1) \Rightarrow r = \frac{3}{4} (2 x + 1)$

$∴ V = \frac{4}{3} π (\frac{3}{4})^{3} (2 x + 1)^{3} = \frac{9}{16} π (2 x + 1)^{3}$

Hence, the rate of change of volume with respect to x is:

$\frac{dV}{dx} = \frac{9}{16} π \frac{d}{dx} (2 x + 1)^{3} = \frac{9}{16} π \times 3 (2 x + 1)^{2} \times 2 = \frac{27}{8} π (2 x + 1)^{2}$

Sand is pouring from a pipe at the rate of $12 cm^{3} / s$ . The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm? Sol. The volume of a cone (V) with radius (r) and height (h) is given by,

$V = \frac{1}{3} π r^{2} h$

It is given that,

$∴ \frac{dV}{dt} = 12 cm^{3} / s and h = \frac{1}{6} r \Rightarrow r = 6 h V = \frac{1}{3} π (6 h)^{2} h = 12 π h^{3}$

The rate of change of volume with respect to time (t) is given by,

$\frac{dV}{dt} \frac{dV}{dt} = 12 π \frac{d}{dh} (h^{3}) \cdot \frac{dh}{dt} = 12 π (3 h^{2}) \frac{dh}{dt} = 36 π h^{2} \frac{dh}{dt} = 12 cm^{3} / s$

Therefore, when $h = 4 cm$ , we have:

$12 = 36 π (4)^{2} \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt} = \frac{12}{36 π ( 16 )} = \frac{1}{48 π} cm / s [\frac{dV}{dt} = 12 cm^{3} / s]$ Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of $\frac{1}{48 π} cm / s$ .

The total cost C (x) in Rupees associated with the production of x units of an item is given by $C (x) = 0.007 x^{3} - 0.003 x^{2} + 15 x + 4000 :$ Find the marginal cost when 17 units are produced.

Sol. Marginal cost is the rate of change of total cost with respect to output.

$Marginal cost (MC) = \frac{d C}{d x} = 0.007 (3 x^{2}) - 0.003 (2 x) + 15 = 0.021 x^{2} - 0.006 x + 15$

When $x = 17$ ,

$MC = 0.021 (1 7^{2}) - 0.006 (17) + 15 = 0.021 (289) - 0.006 (17) + 15 = 6.069 - 0.102 + 15 = 20.967$

Hence, when 17 units are produced, the marginal cost is ₹20.967.

The total revenue in Rupees received from the sale of xv units of a product is given by $R (x) = 13 x^{2} + 26 x + 15$ Find the marginal revenue when $x = 7$ . Sol. Marginal revenue is the rate of change of total revenue with respect to the number of units sold. ∴ Marginal Revenue (MR)

$= \frac{dR}{dx} = 13 (2 x) + 26 = 26 x + 26$

When $x = 7$ ,

$MR = 26 (7) + 26 = 182 + 26 = 208.$

Hence, the required marginal revenue is ₹208.

Choose the correct answer in the following questions from 17 & 18.

The rate of change of the area of a circle with respect to its radius r at $r = 6 cm$ is? (A) $10 π$ (B) $12 π$ (C) $8 π$ (D) $11 π$ Sol. (B) The area of a circle (A) with radius (r) is given by, $A = π r^{2}$ Therefore, the rate of change of the area with respect to its radius r is $\frac{dA}{dr} = \frac{d}{dr} (π r^{2}) = 2 π r$ . $∴$ When $r = 6 cm, \frac{dA}{dr} = 2 π \times 6 = 12 π cm^{2} / cm$ Hence, the required rate of change of the area of a circle is $12 π cm^{2} / cm$ The correct answer is B.
The total revenue in Rupees received from the sale of x units of a product is given by $R (x) = 3 x^{2} + 36 x + 5$ . The marginal revenue, when $x = 15$ , is? (A) 116 (B) 96 (C) 90 (D) 126 Sol. (D) Marginal revenue is the rate of change of total revenue with respect to the number of units sold. ∴ Marginal Revenue (MR)

$= \frac{dR}{dx} = 3 (2 x) + 36 = 6 x + 36$

$∴$ When $x = 15$ ,

$MR = 6 (15) + 36 = 90 + 36 = 126$

Hence, the required marginal revenue is ₹126. The correct answer is D.

EXERCISE - 6.2

Show that the function given by $f (x) = 3 x + 17$ is strictly increasing on R. Sol. Let $x_{1}, x_{2} \in R$ , such that $x_{1} < x_{2} \Rightarrow 3 x_{1} < 3 x_{2}$

$\Rightarrow 3 x_{1} + 17 < 3 x_{2} + 17 \Rightarrow f (x_{1}) < f (x_{2})$

Hence, $f$ is strictly increasing on $R$ . Alternate method: Given $f (x) = 3 x + 17$ . On diff. w.r.t. x. we get $f^{'} (x) = 3 > 0$ , in every interval of R. Thus, the function is strictly increasing on R .

Show that the function given by $f (x) = e^{2 x}$ is strictly increasing on R. Sol. Let $x_{1}, x_{2} \in R$ , such that $x_{1} < x_{2}$ $\Rightarrow 2 x_{1} < 2 x_{2}$ $\Rightarrow e^{2 x_{1}} < e^{2 x_{2}}$ $\Rightarrow f (x_{1}) < f (x_{2})$

Hence, f is strictly increasing on R .

Show that the function given by $f (x) = sin x$ is (a) strictly increasing in $(0, \frac{π}{2})$ (b) strictly decreasing in $(\frac{π}{2}, π)$ (c) neither increasing nor decreasing in $(0, π)$

Sol. The given function is $f (x) = sin x$ .

$∴ f^{'} (x) = cos x$ (a) Since for each $x \in (0, \frac{π}{2}), cos x > 0$ we have $f^{'} (x) > 0$ . Hence, f is strictly increasing in $(0, \frac{π}{2})$ (b) Since for each $x \in (\frac{π}{2}, π), cos x < 0$ , we have $f^{'} (x) < 0$ . Hence, f is strictly decreasing in $(\frac{π}{2}, π)$ . (c) From the results obtained in (a) and (b) it is clear that f is strictly increasing in $(0, \frac{π}{2})$ and strictly decreasing in $(\frac{π}{2}, π)$ so, f is neither increasing nor decreasing in $(0, π)$ .

Find the intervals in which the function f given by $f (x) = 2 x^{2} - 3 x$ is (a) strictly increasing (b) strictly decreasing Sol. The given function is $f (x) = 2 x^{2} - 3 x$ $\Rightarrow f^{'} (x) = 4 x - 3$ (a) For strictly increasing: $f^{'} (x) > 0$ $\Rightarrow 4 x - 3 > 0 \Rightarrow x > \frac{3}{4} \Rightarrow (\frac{3}{4}, \infty)$ (b) For strictly decreasing : $f^{'} (x) < 0$ $\Rightarrow 4 x - 3 < 0 \Rightarrow x < \frac{3}{4} \Rightarrow (- \infty, \frac{3}{4})$
Find the intervals in which the function f given by $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 7$ is (a) strictly increasing (b) strictly decreasing Sol. The given function is

$f (x) f^{'} (x) = 2 x^{3} - 3 x^{2} - 36 x + 7. = 6 x^{2} - 6 x - 36 = 6 (x^{2} - x - 6) = 6 (x + 2) (x - 3)$

(a) For strictly increasing: $f^{'} (x) > 0$ $\Rightarrow 6 (x + 2) (x - 3) > 0$ $\Rightarrow x \in (- \infty, - 2) \cup (3, \infty)$

(b) For strictly decreasing : $f^{'} (x) < 0$ $\Rightarrow 6 (x + 2) (x - 3) < 0$ $\Rightarrow x \in (- 2, 3)$

Hence, the given function $f (x)$ is strictly increasing in intervals $(- \infty, - 2) \cup (3, \infty)$ , while function $f (x)$ is strictly decreasing in interval (-2, 3).

Find the intervals in which the following functions are strictly increasing or decreasing: (a) $x^{2} + 2 x - 5$ (b) $10 - 6 x - 2 x^{2}$ (c) $- 2 x^{3} - 9 x^{2} - 12 x + 1$ (d) $6 - 9 x - x^{2}$ (e) $(x + 1)^{3} (x - 3)^{3}$ Sol. (a) We have, $f (x) = x^{2} + 2 x - 5$ $∴ f^{'} (x) = 2 x + 2$ (i) For strictly increasing: $f^{'} (x) > 0$ $\Rightarrow 2 x + 2 > 0 \Rightarrow x > - 1$ $\Rightarrow x \in (- 1, \infty)$ (ii) For strictly decreasing : $f^{'} (x) < 0$ . $\Rightarrow 2 x + 2 < 0 \Rightarrow x < - 1$ $\Rightarrow x \in (- \infty, - 1)$

(b) We have, $f (x) = 10 - 6 x - 2 x^{2}$

$∴ f^{'} (x) = - 6 - 4 x$

(i) For strictly increasing: $f^{'} (x) > 0$

$\Rightarrow - 6 - 4 x > 0 \Rightarrow x < - \frac{3}{2}$

$\Rightarrow x \in (- \infty, - \frac{3}{2})$

(ii) For strictly decreasing: $f^{'} (x) < 0$

$\Rightarrow - 6 - 4 x < 0 \Rightarrow x > - \frac{3}{2}$

$\Rightarrow x \in (- \frac{3}{2}, \infty)$

$∴ f^{'} (x) = - 6 x^{2} - 18 x - 12 = - 6 (x^{2} + 3 x + 2) = - 6 (x + 1) (x + 2)$

(i) For strictly increasing: $f^{'} (x) > 0$

$\Rightarrow \Rightarrow \Rightarrow - 6 (x + 1) (x + 2) > 0 6 (x + 1) (x + 2) < 0 x \in (- 2, - 1)$

(ii) For strictly decreasing: $f^{'} (x) < 0$

$\Rightarrow \Rightarrow \Rightarrow - 6 (x + 1) (x + 2) < 0 6 (x + 1) (x + 2) > 0 x \in (- \infty, - 2) \cup (- 1, \infty)$

(d) We have, $f (x) = 6 - 9 x - x^{2}$

$∴ f^{'} (x) = - 9 - 2 x$

(i) For strictly increasing: $f^{'} (x) > 0$

$\Rightarrow - 9 - 2 x > 0 \Rightarrow x < - \frac{9}{2} \Rightarrow x \in (- \infty, - \frac{9}{2})$

(ii) For strictly decreasing : $f^{'} (x) < 0$ .

$\Rightarrow - 9 - 2 x < 0 \Rightarrow x > - \frac{9}{2} \Rightarrow x \in (- \frac{9}{2}, \infty)$

(e) We have, $f (x) = (x + 1)^{3} (x - 3)^{3}$

$f^{'} (x) = 3 (x + 1)^{2} (x - 3)^{3} + 3 (x - 3)^{2} (x + 1)^{3} = 3 (x + 1)^{2} (x - 3)^{2} [x - 3 + x + 1] = 3 (x + 1)^{2} (x - 3)^{2} (2 x - 2) = 6 (x + 1)^{2} (x - 3)^{2} (x - 1)$

(i) For strictly increasing: $f^{'} (x) > 0$

$\Rightarrow \Rightarrow 6 (x + 1)^{2} (x - 3)^{2} (x - 1) > 0 x \in (- \infty, - 1) \cup (- 1, 1)$

(ii) For strictly decreasing : $f^{'} (x) < 0$ .

$\Rightarrow 6 (x + 1)^{2} (x - 3)^{2} (x - 1) < 0 \Rightarrow x \in (1, 3) \cup (3, \infty)$

Show that $y = lo g (1 + x) - \frac{2 x}{2 + x}, x > - 1$ , is an increasing function of x throughout its domain. Sol. We have,

$y = lo g (1 + x) - \frac{2 x}{2 + x}$

$∴ \frac{dy}{dx} = \frac{1}{1 + x} - \frac{( 2 + x ) ( 2 ) - 2 x ( 1 )}{( 2 + x ) ^{2}}$

$= \frac{1}{1 + x} - \frac{4}{( 2 + x ) ^{2}} = \frac{x ^{2}}{( 1 + x ) ( 2 + x ) ^{2}}$

When $x \in (- 1, \infty)$ , then $\frac{x ^{2}}{( 2 + x ) ^{2}} > 0$ and $(1 + x) > 0 \Rightarrow \frac{dy}{dx} > 0$ when $x > - 1$ . $∴ y$ is an increasing function of x throughout its domain , i.e. $(x > - 1)$

Find the values of x for which $y = [x (x - 2)]^{2}$ is an increasing function. Sol. We have, $y = [x (x - 2)]^{2} = [x^{2} - 2 x]^{2}$

$∴ \frac{dy}{dx} = 2 (x^{2} - 2 x) (2 x - 2) = 4 x (x - 2) (x - 1)$

For increasing : $\frac{dy}{dx} \geq 0$ $\Rightarrow 4 x (x - 2) (x - 1) > 0$ $\Rightarrow x \in [0, 1] \cup [2, \infty)$

$∴ y$ is increasing in intervals $[0, 1] \cup [2, \infty)$ .

Prove that $y = \frac{4 s i n θ}{( 2 + c o s θ )} - θ$ is an increasing function of $θ$ in $[0, \frac{π}{2}]$ .

Sol. We have, $y = \frac{4 s i n θ}{( 2 + c o s θ )} - θ$

$\Rightarrow \frac{dy}{d θ} = \frac{( 2 + cos θ ) ( 4 cos θ ) - 4 sin θ ( - sin θ )}{( 2 + cos θ ) ^{2}} - 1 = \frac{8 cos θ + 4 cos ^{2} θ + 4 sin ^{2} θ}{( 2 + cos θ ) ^{2}} - 1 = \frac{8 cos θ + 4}{( 2 + cos θ ) ^{2}} - 1$

$\Rightarrow \frac{dy}{d θ} = \frac{8 c o s θ + 4 - ( 4 + c o s ^{2} θ + 4 c o s θ )}{( 2 + c o s θ ) ^{2}}$

$= \frac{4 cos θ - cos ^{2} θ}{( 2 + cos θ ) ^{2}} = \frac{cos θ ( 4 - cos θ )}{( 2 + cos θ ) ^{2}}$

In interval $[0, \frac{π}{2}]$ , we have $cos θ \geq 0$ . Also, $4 > cos θ \Rightarrow 4 - cos θ > 0$ . $∴ cos θ (4 - cos θ) \geq 0$ and also $(2 + cos θ)^{2} > 0$ $\Rightarrow \frac{c o s θ ( 4 - c o s θ )}{( 2 + c o s θ ) ^{2}} \geq 0$ $\Rightarrow \frac{dy}{dx} \geq 0$ Therefore, y is increasing in interval $[0, \frac{π}{2}]$ .

Prove that the logarithmic function is strictly increasing on $(0, \infty)$ . Sol. Let $f (x) = lo g x . ∴ f^{'} (x) = \frac{1}{x}$ It is clear that for $x > 0, f^{'} (x) = \frac{1}{x} > 0$ Hence, $f (x) = lo g x$ is strictly increasing in interval $(0, \infty)$ .
Prove that the function f given by $f (x) = x^{2} - x + 1$ is neither strictly increasing nor strictly decreasing on (-1, 1). Sol. The given function is $f (x) = x^{2} - x + 1$ and $x \in (- 1, 1)$ . $∴ f^{'} (x) = 2 x - 1$ For strictly increasing: $f^{'} (x) > 0$ $\Rightarrow 2 x - 1 > 0 \Rightarrow x > \frac{1}{2} \Rightarrow x \in (\frac{1}{2}, 1)$ For strictly decreasing : $f^{'} (x) < 0$ $\Rightarrow 2 x - 1 < 0 \Rightarrow x < \frac{1}{2} \Rightarrow x \in (- 1, \frac{1}{2})$ Since, $f (x)$ is strictly increasing in $(\frac{1}{2}, 1)$ and strictly decreasing $(- 1, \frac{1}{2})$ . Hence, f is neither strictly increasing nor strictly decreasing in interval (-1, 1).
Which of the following functions are strictly decreasing on $(0, \frac{π}{2})$ ? (A) $cos x$ (B) $cos 2 x$ (C) $cos 3 x$ (D) $tan x$

Sol. (A) Let $f_{1} (x) = cos x$ $∴ f_{1}^{'} (x) = - sin x$ In interval $(0, \frac{π}{2}), f_{1}^{'} (x) = - sin x < 0$ $∴ f_{1} (x)$ is strictly decreasing in interval $(0, \frac{π}{2})$ .

(B) Let $f_{2} (x) = cos 2 x$ $∴ f_{2}^{'} (x) = - 2 sin 2 x$ Now, $0 < x < \frac{π}{2}$ $\Rightarrow 0 < 2 x < π \Rightarrow sin 2 x > 0 \Rightarrow - 2 sin 2 x < 0$ $∴ f_{2}^{'} (x) = - 2 sin 2 x < 0$ on $(0, \frac{π}{2})$ $∴ f_{2} (x)$ is strictly decreasing in interval $(0, \frac{π}{2})$ . (C) Let $f_{3} (x) = cos 3 x$ . $∴ f_{3}^{'} (x) = - 3 sin 3 x$ Now, $f_{3}^{'} (x) = 0$ $\Rightarrow sin 3 x = 0 \Rightarrow 3 x = π$ , as $x \in (0, \frac{π}{2}) \Rightarrow x = \frac{π}{3}$ The point $x = \frac{π}{3}$ divides the interval $(0, \frac{π}{2})$ into two disjoint intervals i.e., $(0, \frac{π}{3})$ and $(\frac{π}{3}, \frac{π}{2})$ Now in interval $(0, \frac{π}{3}), f_{3}^{'} (x) = - 3 sin 3 x < 0$

$[as 0 < x < \frac{π}{3} \Rightarrow 0 < 3 x < π]$

$∴ f_{3} (x)$ is strictly decreasing in interval $(0, \frac{π}{3})$ However, in interval $(\frac{π}{3}, \frac{π}{2})$ , $f_{3}^{'} (x) = - 3 sin 3 x > 0$

$[as \frac{π}{3} < x < \frac{π}{2} \Rightarrow π < 3 x < \frac{3 π}{2}]$

$∴ f_{3} (x)$ is strictly increasing in interval $(\frac{π}{3}, \frac{π}{2})$ . Hence, $f_{3}$ is neither strictly increasing nor strictly decreasing in interval $(0, \frac{π}{2})$ . (D) Let $f_{4} (x) = tan x$ . $∴ f_{4}^{'} (x) = sec^{2} x$ In interval $(0, \frac{π}{2}), f_{4}^{'} (x) = sec^{2} x > 0$

$∴ f_{4}$ is strictly increasing in interval $(0, \frac{π}{2})$ Therefore, functions $cos x$ and $cos 2 x$ are strictly decreasing in $(0, \frac{π}{2})$ Hence, the correct answers are A and B.

On which of the following intervals is the function f given by $f (x) = x^{100} + sin x - 1$ strictly decreasing? (A) $(0, 1)$ (B) $(\frac{π}{2}, π)$ (C) $(0, \frac{π}{2})$ (D) None of these

Sol. (D) We have, $f (x) = x^{100} + sin x - 1$

$∴ f^{'} (x) = 100 x^{99} + cos x$ In interval (0, 1), $cos x > 0$ and $100 x^{99} > 0$ $∴ f^{'} (x) > 0$ Thus, function f is strictly increasing in interval (0, 1). In interval $(\frac{π}{2}, π), cos x < 0$ and $100 x^{99} > 0$ . Also, $100 x^{99} > cos x$ $∴ f^{'} (x) > 0$ in $(\frac{π}{2}, π)$ . Thus, function f is strictly increasing in interval $(\frac{π}{2}, π)$ In interval $(0, \frac{π}{2}), cos x > 0$ and $100 x^{99} > 0$ $∴ 100 x^{99} + cos x > 0$ $\Rightarrow f^{'} (x) > 0$ on $(0, \frac{π}{2})$ $∴ f$ is strictly increasing in interval $(0, \frac{π}{2})$ Hence function f is strictly decreasing in none of the intervals. The correct answer is D.

For what values of a the function f given by $f (x) = x^{2} + ax + 1$ is increasing on [1,2]?

Sol. We have, $f (x) = x^{2} + ax + 1$

$∴ f^{'} (x) = 2 x + a$ Now, $1 \leq x \leq 2$ $\Rightarrow 2 + a \leq 2 x + a \leq 4 + a$ $\Rightarrow 2 + a \leq f^{'} (x) \leq 4 + a$

Now, function f will be increasing in [1, 2], if $f^{'} (x) \geq 0$ .

$\Rightarrow 2 + a \geq 0 \Rightarrow a \geq - 2$ and $4 + a \geq 0 \Rightarrow a \geq - 4$ $[- 2, \infty) \cap [- 4, \infty)$ $a \in [- 2, \infty)$ $a \geq - 2$

Let I be any interval disjoint from [-1, 1]. Prove that the function f given by $f (x) = x + \frac{1}{x}$ is strictly increasing on I. Sol. We have, $f (x) = x + \frac{1}{x}$

$∴ f^{'} (x) = 1 - \frac{1}{x ^{2}}$ $\Rightarrow f^{'} (x) = \frac{x ^{2} - 1}{x ^{2}} = \frac{( x + 1 ) ( x - 1 )}{x ^{2}}$

For strictly increasing: $f^{'} (x) > 0$

$\Rightarrow \frac{( x + 1 ) ( x - 1 )}{x ^{2}} > 0 \Rightarrow x \in (- \infty, - 1) \cup (1, \infty)$ $∴ f$ is strictly increasing on $(- \infty, - 1) \cup (1, \infty)$

Hence, function f is strictly increasing in interval I disjoint from [-1, 1]. Hence, the given result is proved.

Prove that the function f given by $f (x) = lo g sin x$ is strictly increasing on $(0, \frac{π}{2})$ and strictly decreasing on $(\frac{π}{2}, π)$ . Sol. We have, $f (x) = lo g sin x$ $∴ f^{'} (x) = \frac{1}{s i n x} (cos x) = cot x$ In interval $(0, \frac{π}{2})$ ,

$f^{'} (x) = cot x > 0$

$∴ f$ is strictly increasing on $(0, \frac{π}{2})$ . In interval $(\frac{π}{2}, π)$ ,

$f^{'} (x) = cot x < 0.$

$∴ f$ is strictly decreasing on $(\frac{π}{2}, π)$ .

Prove that the function $f$ given by $f (x) = lo g cos x$ is strictly decreasing on $(0, \frac{π}{2})$ and strictly increasing on $(\frac{π}{2}, π)$ . Sol. We have, $f (x) = lo g cos x$ $∴ f^{'} (x) = \frac{1}{c o s x} (- sin x) = - tan x$ In interval $(0, \frac{π}{2})$ , $tan x > 0 \Rightarrow - tan x < 0$ $∴ f^{'} (x) < 0$ on $(0, \frac{π}{2})$ $∴ f$ is strictly decreasing on $(0, \frac{π}{2})$ . In interval $(\frac{π}{2}, π)$ , $tan x < 0 \Rightarrow - tan x > 0$ $∴ f^{'} (x) > 0$ on $(\frac{π}{2}, π)$ $∴ f$ is strictly increasing on $(\frac{π}{2}, π)$
Prove that the function given by $f (x) = x^{3} - 3 x^{2} + 3 x - 100$ is increasing in $R$ . Sol. We have, $f (x) = x^{3} - 3 x^{2} + 3 x - 100$

$f^{'} (x) = 3 x^{2} - 6 x + 3 = 3 (x^{2} - 2 x + 1) = 3 (x - 1)^{2}$

For any $x \in R, (x - 1)^{2} \geq 0$ . $\Rightarrow 3 (x - 1)^{2} \geq 0$ $\Rightarrow f^{'} (x) \geq 0\forall x \in R$

Hence, the given function $f (x)$ is increasing in R.

The interval in which the function $f$ given by $f (x) = x^{2} e^{- x}$ is strictly increasing, is (A) $(- \infty, \infty)$ (B) $(- \infty, 0)$ (C) $(2, \infty)$ (D) $(0, 2)$ Sol. (D) $f (x) = x^{2} . e^{- x}$ (Differentiating w.r.t. x)

$f^{'} (x) = - x^{2} e^{- x} + 2 x e^{- x}$

put $f^{'} (x) > 0 - x^{2} . e^{- x} + 2 x . e^{- x} > 0$ $\Rightarrow e^{- x} (2 x - x^{2}) > 0$ $\Rightarrow e^{- x} x (2 - x) > 0$ $\Rightarrow e^{- x} x (x - 2) < 0$ ( $∵ e^{- x} \neq = 0$ , as $e^{- x}$ is always positive for all $x \in R$ ) when $x \in (0, 2), f^{'} (x) > 0$ so f(x) is strictly increasing function in interval $(0, 2)$

EXERCISE - 6.3

Find the maximum and minimum values, if any, of the following functions given by: (i) $f (x) = (2 x - 1)^{2} + 3$ (ii) $f (x) = 9 x^{2} + 12 x + 2$ (iii) $f (x) = - (x - 1)^{2} + 10$ (iv) $g (x) = x^{3} + 1$

Sol.

(i) The given function is $f (x) = (2 x - 1)^{2} + 3$ . It can be observed that $(2 x - 1)^{2} \geq 0$ for every $x \in R$ . Therefore, $f (x) = (2 x - 1)^{2} + 3 \geq 3$ for every $x \in R$ . $∴$ Minimum value of $f (x) = 3$ and function $f (x)$ does not have a maximum value. (ii) The given function is $f (x) = 9 x^{2} + 12 x + 2 =$ $(3 x + 2)^{2} - 2$ . It can be observed that $(3 x + 2)^{2} \geq 0$ for every $x \in R$ . Therefore, $f (x) = (3 x + 2)^{2} - 2 \geq - 2$ for every $x \in R$ . ∴ Minimum value of $f (x) = - 2$ and function f does not have a maximum value. (iii) The given function is $f (x) = - (x - 1)^{2} + 10$ . It can be observed that $(x - 1)^{2} \geq 0$ for every $x \in R$ . $\Rightarrow - (x - 1)^{2} \leq 0$ for every $x \in R$

Therefore, $f (x) = - (x - 1)^{2} + 10 \leq 10$ for every $x \in R$ .

∴ Maximum value of $f (x) = 10$ and function $f (x)$ does not have a minimum value. (iv) The given function is $g (x) = x^{3} + 1$ . At $x \to \infty g (x) \to \infty$ At $x \to - \infty g (x) \to - \infty$ Hence, function $g (x)$ neither has a maximum value nor a minimum value.

Find the maximum and minimum values, if any, of the following functions given by: (i) $f (x) = ∣ x + 2∣ - 1$ (ii) $g (x) = - ∣ x + 1∣ + 3$ (iii) $h (x) = sin (2 x) + 5$ (iv) $f (x) = ∣ sin 4 x + 3∣$ (v) $h (x) = x + 1, x \in (- 1, 1)$

Sol.

(i) $f (x) = ∣ x + 2∣ - 1$ We know that $∣ x + 2∣ \geq 0$ for every $x \in R$ . Therefore, $f (x) = ∣ x + 2∣ - 1 \geq - 1$ for every $x \in R$ . $∴$ Minimum value of $f (x) = - 1$ and function $f (x)$ does not have a maximum value.

(ii) $g (x) = - ∣ x + 1∣ + 3$ We know that $- ∣ x + 1∣ \leq 0$ for every $x \in R$ . Therefore, $g (x) = - ∣ x + 1∣ + 3 \leq 3$ for every $x \in R$ . $∴$ Maximum value of $g (x) = 3$ and function $g (x)$ does not have a minimum value. (iii) $h (x) = sin 2 x + 5$ We know that $- 1 \leq sin 2 x \leq 1$ . $∴ - 1 + 5 \leq sin 2 x + 5 \leq 1 + 5$ $∴ 4 \leq sin 2 x + 5 \leq 6 \Rightarrow 4 \leq h (x) \leq 6$ Hence, the maximum and minimum values of $h (x)$ are 6 and 4 respectively. (iv) $f (x) = ∣ sin 4 x + 3∣$ We know that $- 1 \leq sin 4 x \leq 1$ . $∴ 2 \leq sin 4 x + 3 \leq 4$ $∴ 2 \leq ∣ sin 4 x + 3∣ \leq 4 \Rightarrow 2 \leq f (x) \leq 4$ Hence, the maximum and minimum values of $f (x)$ are 4 and 2 respectively. (v) $h (x) = x + 1, x \in (- 1, 1)$ as $- 1 < x < 1$ $\Rightarrow - 1 + 1 < x + 1 < 1 + 1 \Rightarrow 0 < h (x) < 2$ Function $h (x)$ has neither maximum nor minimum value in (-1, 1).

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i) $f (x) = x^{2}$ (ii) $g (x) = x^{3} - 3 x$ (iii) $h (x) = sin x + cos x, 0 < x < \frac{π}{2}$ (iv) $f (x) = sin x - cos x, 0 < x < 2 π$ (v) $f (x) = x^{3} - 6 x^{2} + 9 x + 15$ (vi) $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$ (vii) $g (x) = \frac{1}{x ^{2} + 2}$ (viii) $f (x) = x 1 - x, x > 0$ Sol. (i) $f (x) = x^{2}$ $\Rightarrow f^{'} (x) = 2 x$ and $f^{''} (x) = 2$ Now for maxima or minima $f^{'} (x) = 0$ $2 x = 0 \Rightarrow x = 0$ at $x = 0 f^{''} (x) = 2 > 0$ Therefore, by second derivative test, $x = 0$ is a point of local minima and local minimum value of $f (x)$ at $x = 0$ is $f (0) = 0$ . (ii) $g (x) = x^{3} - 3 x$ $\Rightarrow g^{'} (x) = 3 x^{2} - 3$ and $g^{''} (x) = 6 x$ Now for maxima or minima $g^{'} (x) = 0$ $\Rightarrow 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ when $x = 1 g^{''} (1) = 6 > 0$ when $x = - 1 g^{''} (- 1) = - 6 < 0$ By second derivative test, $x = 1$ is a point of local minima and local minimum value of $g (x)$ at $x = 1$ is $g (1) = 1^{3} - 3 = 1 - 3 = - 2$ . However, $x = - 1$ is a point of local maxima and local maximum value of $g (x)$ at $x = - 1$ is $g (- 1) = (- 1)^{3} - 3 (- 1) = - 1 + 3 = 2$ . (iii) $h (x) = sin x + cos x, 0 < x < \frac{π}{2}$ $\Rightarrow h^{'} (x) = cos x - sin x$ for maxima or minima $h^{'} (x) = 0 \Rightarrow sin x = cos x \Rightarrow tan x = 1$ $\Rightarrow x = \frac{π}{4} \in (0, \frac{π}{2})$ h' (x) = - sin x - cos x = $- (sin x + cos x)$ $h^{''} (\frac{π}{4}) = - (\frac{1}{2} + \frac{1}{2}) = - \frac{2}{2} = - 2 < 0$ Therefore, by second derivative test, $x = \frac{π}{4}$ is a point of local maxima and the local maximum value of $h (x)$ at $x = \frac{π}{4}$ is $h (\frac{π}{4}) = sin \frac{π}{4} + cos \frac{π}{4} = \frac{1}{2} + \frac{1}{2} = 2$

(iv) $f (x) = sin x - cos x, 0 < x < 2 π$

$∴ f^{'} (x) = cos x + sin x$

for maxima or minima

$f^{'} (x) = 0 \Rightarrow cos x = - sin x \Rightarrow tan x = - 1$

$\Rightarrow x = \frac{3 π}{4}, \frac{7 π}{4} \in (0, 2 π)$

$f^{''} (x) = - sin x + cos x$

$f^{''} (\frac{3 π}{4}) = - sin \frac{3 π}{4} + cos \frac{3 π}{4} = - \frac{1}{2} - \frac{1}{2} = - 2 < 0$

$f^{''} (\frac{7 π}{4}) = - sin \frac{7 π}{4} + cos \frac{7 π}{4} = \frac{1}{2} + \frac{1}{2} = 2 > 0$

Therefore, by second derivative test, $x = \frac{3 π}{4}$ is a point of local maxima and the local maximum value of $f (x)$ at $x = \frac{3 π}{4}$ is $f (\frac{3 π}{4}) = sin \frac{3 π}{4} - cos \frac{3 π}{4} = \frac{1}{2} + \frac{1}{2} = 2$ . However, $x = \frac{7 π}{4}$ is a point of local minima and the local minimum value of $f (x)$ at $x = \frac{7 π}{4}$ is $f (\frac{7 π}{4}) = sin \frac{7 x}{4} - cos \frac{7 π}{4} = - \frac{1}{2} - \frac{1}{2} = - 2$ (v) $f (x) = x^{3} - 6 x^{2} + 9 x + 15$ and $f^{'} (x) = 3 x^{2} - 12 x + 9$ for maxima or minima

$f^{'} (x) = 0 \Rightarrow 3 (x^{2} - 4 x + 3) = 0$

$\Rightarrow 3 (x - 1) (x - 3) = 0 \Rightarrow x = 1, 3$ Now, $f^{''} (x) = 6 x - 12 = 6 (x - 2)$ when $x = 1, f^{''} (1) = 6 (1 - 2) = - 6 < 0$ when $x = 3, f^{''} (3) = 6 (3 - 2) = 6 > 0$ Therefore, by second derivative test, $x = 1$ is a point of local maxima and the local maximum value of $f (x)$ at $x = 1$ is

$f (1) = 1 - 6 + 9 + 15 = 19.$

However, $x = 3$ is a point of local minima and the local minimum value of $f (x)$ at $x = 3$ is

$f (3) = 27 - 54 + 27 + 15 = 15.$

(vi) $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$ and $g^{'} (x) = \frac{1}{2} - \frac{2}{x ^{2}}$ for maxima or minima

$g^{'} (x) = 0 gives \frac{2}{x ^{2}} = \frac{1}{2} \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

Since $x > 0$ , we take $x = 2$ . Now, $g^{''} (x) = \frac{4}{x ^{3}} \Rightarrow g^{''} (2) = \frac{4}{2 ^{3}} = \frac{1}{2} > 0$ Therefore, by second derivative test, $x = 2$ is a point of local minima and the local minimum value of $g (x)$ at $x = 2$ is $g (2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$ (vii) $g (x) = \frac{1}{x ^{2} + 2}$ and $g^{'} (x) = \frac{- ( 2 x )}{( x ^{2} + 2 ) ^{2}}$ for maxima or minima,

$g^{'} (x) = 0 \Rightarrow \frac{- 2 x}{( x ^{2} + 2 ) ^{2}} = 0 \Rightarrow x = 0$

Now, for values close to $x = 0$ and to the left of 0, $g^{'} (x) > 0$ . Also, for values close to $x = 0$ and to the right of 0, $g^{'} (x) < 0$ . Therefore, by first derivative test, $x = 0$ is a point of local maxima and the local maximum value of $g (0)$ is $\frac{1}{0 + 2} = \frac{1}{2}$ (viii) $f (x) = x 1 - x, 0 < x < 1$

$∴ f^{'} (x) = 1 - x + x \cdot \frac{1}{2 1 - x} (- 1) = 1 - x - \frac{x}{2 1 - x} = \frac{2 ( 1 - x ) - x}{2 1 - x} = \frac{2 - 3 x}{2 1 - x}$

for maxima or minima

$\Rightarrow f^{'} (x) = 0 \Rightarrow \frac{2 - 3 x}{2 1 - x} = 0^{'}$

$\Rightarrow 2 - 3 x = 0 \Rightarrow x = \frac{2}{3}$

$\Rightarrow f^{'} (x) = \frac{2 - 3 x}{2 1 - x}$

$f^{''} (x) f^{''} (\frac{2}{3}) = \frac{1}{2} \frac{1 - x ( - 3 ) - ( 2 - 3 x ) ( \frac{- 1}{2 1 - x} )}{1 - x} = \frac{1 - x ( - 3 ) + ( 2 - 3 x ) ( \frac{1}{2 1 - x} )}{2 ( 1 - x )} = \frac{- 6 ( 1 - x ) + ( 2 - 3 x )}{4 ( 1 - x ) ^{\frac{3}{2}}} = \frac{3 x - 4}{4 ( 1 - x ) ^{\frac{3}{2}}} = \frac{3 ( \frac{2}{3} ) - 4}{4 ( 1 - \frac{2}{3} ) ^{\frac{3}{2}}} = \frac{2 - 4}{4 ( \frac{1}{3} ) ^{\frac{3}{2}}} = \frac{- 1}{2 ( \frac{1}{3} ) ^{\frac{3}{2}}} < 0$

Therefore, by second derivative test, $x = \frac{2}{3}$ is a point of local maxima and the local maximum value of $f (x)$ at $x = \frac{2}{3}$ is

$f (\frac{2}{3}) = \frac{2}{3} 1 - \frac{2}{3} = \frac{2}{3} \frac{1}{3} = \frac{2}{3 3} = \frac{2 3}{9}$

Prove that the following functions do not have maxima or minima: (i) $f (x) = e^{x}$ (ii) $g (x) = lo g x$ (iii) $h (x) = x^{3} + x^{2} + x + 1$ Sol. (i) We have, $f (x) = e^{x}$ $∴ f^{'} (x) = e^{x}$ Now, if $f^{'} (x) = 0$ , then $e^{x} = 0$ . But, the exponential function can never assume 0 for any value of x. Therefore, there does not exist $x \in R$ such that $f^{'} (x) = 0$ Hence, function f does not have maxima or minima.

(ii) We have, $g (x) = lo g x$ $∴ g^{'} (x) = \frac{1}{x}$ Since $lo g x$ is defined for a positive number x , $g^{'} (x) > 0$ for any x. Therefore, there does not exist $x \in R$ such that $g^{'} (x) = 0$ . Hence, function g does not have maxima or minima. (iii) We have, $h (x) = x^{3} + x^{2} + x + 1$ $∴ h^{'} (x) = 3 x^{2} + 2 x + 1$ Now, $h^{'} (x) = 0$ $\Rightarrow 3 x^{2} + 2 x + 1 = 0$ $\Rightarrow x = \frac{- 2 \pm 2 2 i}{6} = \frac{- 1 \pm 2 i}{3} \in / R$ Therefore, there does not exist $x \in R$ such that $h^{'} (x) = 0$ . Hence, function h does not have maxima or minima.

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) $f (x) = x^{3}, x \in [- 2, 2]$ (ii) $f (x) = sin x + cos x, x \in [0, π]$ (iii) $f (x) = 4 x - \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$ (iv) $f (x) = (x - 1)^{2} + 3, x \in [- 3, 1]$

Sol.

(i) The given function is $f (x) = x^{3}$ . $∴ f^{'} (x) = 3 x^{2}$ Now, for maxima or minima Put $f^{'} (x) = 0 \Rightarrow x = 0$ Then, we evaluate the value of f(x) at critical point $x = 0$ and at end points of the interval [-2, 2]. At $x = 0, f (0) = 0$ At $x = - 2, f (- 2) = (- 2)^{3} = - 8$

At $x = 2, f (2) = (2)^{3} = 8$ Hence, we can conclude that the absolute maximum value of $f (x)$ on [-2, 2] is 8 occurring at $x = 2$ . Also, the absolute minimum value of f(x) on [-2, 2] is -8 occurring at $x = - 2$ .

(ii) The given function is $f (x) = sin x + cos x$ . $∴ f^{'} (x) = cos x - sin x$ Now, for maxima or minima $f^{'} (x) = 0$ $\Rightarrow sin x = cos x \Rightarrow tan x = 1 \Rightarrow x = \frac{π}{4}$

Then, we evaluate the value of f(x) at critical point $x = \frac{π}{4}$ and at the end points of the interval $[0, π]$ .

$f (\frac{π}{4}) = sin \frac{π}{4} + cos \frac{π}{4} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 2 f (0) = sin 0 + cos 0 = 0 + 1 = 1 f (π) = sin π + cos π = 0 - 1 = - 1$

Hence, we can conclude that the absolute maximum value of $f (x)$ on $[0, π]$ is $2$ occurring at $x = \frac{π}{4}$ and the absolute minimum value of $f (x)$ on $[0, π]$ is -1 occurring at $x = π$ .

(iii) The given function is $f (x) = 4 x - \frac{1}{2} x^{2}$ and

$f^{'} (x) = 4 - \frac{1}{2} (2 x) = 4 - x$

Now, for maxima or minima $f^{'} (x) = 0 \Rightarrow x = 4$ Then, we evaluate the value of f(x) at critical point $x = 4$ and at the end points of the interval

$[- 2, \frac{9}{2}] .$

$f (4) = 16 - \frac{1}{2} (16) = 16 - 8 = 8 f (- 2) = - 8 - \frac{1}{2} (4) = - 8 - 2 = - 10$

$f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} (\frac{9}{2})^{2} = 18 - \frac{81}{8} = 18 - 10.125 = 7.875$

Hence, we can conclude that the absolute maximum value of $f (x)$ on $[- 2, \frac{9}{2}]$ is 8 occurring at $x = 4$ and the absolute minimum value of $f (x)$ on $[- 2, \frac{9}{2}]$ is -10 occurring at $x = - 2$ . (iv) The given function is $f (x) = (x - 1)^{2} + 3$ and $f^{'} (x) = 2 (x - 1)$ Now, for maxima or minima

$f^{'} (x) = 0 \Rightarrow 2 (x - 1) = 0 \Rightarrow x = 1$

Then, we evaluate the value of f(x) at critical point $x = 1$ and at the end points of the interval [-3, 1].

$f (1) = (1 - 1)^{2} + 3 = 0 + 3 = 3 f (- 3) = (- 3 - 1)^{2} + 3 = 16 + 3 = 19$

Hence, we can conclude that the absolute maximum value of $f (x)$ on $[- 3, 1]$ is 19 occurring at $x = - 3$ and the minimum value of $f (x)$ on $[- 3, 1]$ is 3 occurring at $x = 1$ .

Find the maximum profit that a company can make, if the profit function is given by $p (x) = 41 - 72 x - 18 x^{2}$ . Sol. The profit function is,

$p (x) = 41 - 72 x - 18 x^{2}$

$\Rightarrow p^{'} (x) = - 72 - 36 x and p^{''} (x) = - 36$

for maxima or minima

$p^{'} (x) = 0$

$\Rightarrow \Rightarrow - 72 - 36 x = 0 x = \frac{- 72}{36} \Rightarrow x = - 2 p^{''} (- 2) = - 36 < 0$

Then, by second derivative test, $x = - 2$ is point of local maxima of p(x). So, the local maximum value is

$p (- 2) = 41 - 72 (- 2) - 18 (- 2)^{2} = 41 + 144 - 72 = 113$

Therefore, the maximum profit that the company can make is 113 units.

Find both the maximum value and minimum value of $3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25$ on the interval [0, 3].

Sol. Let $f (x) = 3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25$ on $[0, 3]$

$∴ f^{'} (x) = 12 x^{3} - 24 x^{2} + 24 x - 48$ for maxima or minima $f^{'} (x) = 0$ $\Rightarrow 12 x^{3} - 24 x^{2} + 24 x - 48 = 0$ $\Rightarrow x^{3} - 2 x^{2} + 2 x - 4 = 0$ $\Rightarrow (x - 2) (x^{2} + 2) = 0$ $\Rightarrow x = 2 (∵ x^{2} + 2 \neq = 0)$ $∴ x = 2$ is turning point. Now we evaluate the value of $f (x)$ at $x = 2$ and at the end points of the interval [0, 3] $f (2) = 3 (16) - 8 (8) + 12 (4) - 48 (2) + 25 = - 39$ $f (0) = 25$ $f (3) = 3 (81) - 8 (27) + 12 (9) - 48 (3) + 25 = 16$ Therefore, absolute minimum value is -39 at $x = 2$ and absolute maximum value is 25 at $x = 0$ .

At what points in the interval $[0, 2 π]$ , does the function $sin 2 x$ attain its maximum value?

Sol. Let $f (x) = sin 2 x$ .

$∴ f^{'} (x) = 2 cos 2 x$

Now, for maxima or minima

$f^{'} (x) = 0 \Rightarrow cos 2 x = 0$

$\Rightarrow 2 x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}$ $\Rightarrow x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ Then, we evaluate the values of f(x) at critical points $x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ , and at the end points of the interval $[0, 2 π]$ .

$f (\frac{π}{4}) = sin \frac{π}{2} = 1, f (\frac{5 π}{4}) = sin \frac{5 π}{2} = 1, f (0) = sin 0 = 0, f (\frac{3 π}{4}) = sin \frac{3 π}{2} = - 1, f (\frac{7 π}{4}) = sin \frac{7 π}{2} = - 1, f (2 π) = sin 2 π = 0$

Hence, we can conclude that the absolute maximum value of $f (x)$ on $[0, 2 π]$ is 1 occurring at $x = \frac{π}{4}$ and $x = \frac{5 π}{4}$

What is the maximum value of the function sin $x + cos x$ ?

Sol. Let $f (x) = sin x + cos x$ .

$∴ f^{'} (x) = cos x - sin x$ for maxima or minima

$f^{'} (x) = 0 \Rightarrow sin x = cos x$

$\Rightarrow tan x = 1 \Rightarrow x = \frac{π}{4}, \frac{5 π}{4} \dots .$ ,

$f^{''} (x) = - sin x - cos x = - (sin x + cos x)$

Now, $f^{''} (x)$ will be negative when $(sin x + cos x)$ is positive i.e., when $sin x$ and $cos x$ are both positive. Also, we know that $sin x$ and $cos x$ both are positive in the first quadrant. Then, $f^{'} (x)$ will be negative when $x \in (0, \frac{π}{2})$ Thus, we consider $x = \frac{π}{4}$

$f^{''} (\frac{π}{4}) = - (sin \frac{π}{4} + cos \frac{π}{4}) = - (\frac{2}{2}) = - 2 < 0$

∴ By second derivative test, f(x) will be the maximum at $x = \frac{π}{4}$ and the maximum value of f(x) is

$f (\frac{π}{4}) = sin \frac{π}{4} + cos \frac{π}{4} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 2$

Find the maximum value of $2 x^{3} - 24 x + 107$ in the interval [1, 3]. Find the maximum value of the same function in $[- 3, - 1]$ .

Sol. Let $f (x) = 2 x^{3} - 24 x + 107$ .

$∴ f^{'} (x) = 6 x^{2} - 24 = 6 (x^{2} - 4)$

Now, for maxima or minima

$f^{'} (x) = 0$

$\Rightarrow 6 (x^{2} - 4) = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

We first consider the interval [1, 3]. Then, we evaluate the value of f(x) at the critical point $x = 2 \in [1, 3]$ and at the end points of the interval [1, 3].

$f (1) = 2 (1) - 24 (1) + 107 = 2 - 24 + 107 = 85 f (2) = 2 (8) - 24 (2) + 107 = 16 - 48 + 107 = 75 f (3) = 2 (27) - 24 (3) + 107 = 54 - 72 + 107 = 89$

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at $x = 3$ . Next, we consider the interval $[- 3, - 1]$ . Now, we evaluate the value of $f (x)$ at the critical point $x = - 2 \in [- 3, - 1]$ and at the end points of the interval $[- 3, - 1]$ .

$f (- 1) = 2 (- 1) - 24 (- 1) + 107 = - 2 + 24 + 107 = 129$

$f (- 3) = 2 (- 27) - 24 (- 3) + 107 = - 54 + 72 + 107 = 125$

Hence, the absolute maximum value of f(x) in the interval $[- 3, - 1]$ is 139 occurring at $x = - 2$ .

It is given that at $x = 1$ , the function $x^{4} - 62 x^{2} + a x + 9$ attains its maximum value, on the interval [0, 2]. Find the value of a. Sol. Let $f (x) = x^{4} - 62 x^{2} + ax + 9$ .

$∴ f^{'} (x) = 4 x^{3} - 124 x + a$

It is given that function f attains its maximum value on the interval [0, 2] at $x = 1$ .

$∴ f^{'} (1) = 0$ $\Rightarrow 4 - 124 + a = 0 \Rightarrow a = 120$ .

Hence, the value of a is 120.

Find the maximum and minimum values of $x + sin 2 x$ on $[0, 2 π]$ . Sol. Let $f (x) = x + sin 2 x$ and $f^{'} (x) = 1 + 2 cos 2 x$ Now, for maxima or minima

$f^{'} (x) = 0$

$\Rightarrow cos 2 x = - \frac{1}{2} = - cos \frac{π}{3} = cos (π - \frac{π}{3}) = cos \frac{2 π}{3} \Rightarrow 2 x = 2 π \pm \frac{2 π}{3}, n \in Z \Rightarrow x = nπ \pm \frac{π}{3}, n \in Z \Rightarrow x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3} \in [0, 2 π]$

Then, we evaluate the value of f(x) at critical points $x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$ and at the end points of the interval $[0, 2 π]$ .

$f (\frac{π}{3}) = \frac{π}{3} + sin \frac{2 π}{3} = \frac{π}{3} + \frac{3}{2} f (\frac{2 π}{3}) = \frac{2 π}{3} + sin \frac{4 π}{3} = \frac{2 π}{3} - \frac{3}{2} f (\frac{4 π}{3}) = \frac{4 π}{3} + sin \frac{8 π}{3} = \frac{4 π}{3} + \frac{3}{2} f (\frac{5 π}{3}) = \frac{5 π}{3} + sin \frac{10 π}{3} = \frac{5 π}{3} - \frac{3}{2} f (0) = 0 + sin 0 = 0 f (2 π) = 2 π + sin 4 π = 2 π + 0 = 2 π + 0 = 2 π$

Hence, we can conclude that the absolute maximum value of $f (x)$ in the interval $[0, 2 π]$ is $2 π$ occurring at $x = 2 π$ and the absolute minimum value of $f (x)$ in the interval $[0, 2 π]$ is 0 occurring at $x = 0$ .

Find two numbers whose sum is 24 and whose product is as large as possible. Sol. Let the two numbers be x and y. According to the question, $x + y = 24$

$\Rightarrow y = 24 - x$

And let z is the product of x and y.

$\Rightarrow \Rightarrow \Rightarrow z = xy z = x (24 - x) z = 24 x - x^{2} \Rightarrow \frac{dz}{dx} = 24 - 2 x and \frac{d ^{2} z}{dx ^{2}} = - 2$

Now to find turning point, $\frac{dz}{dx} = 0$

$\Rightarrow 24 - 2 x = 0 \Rightarrow x = 12 At x = 12, \frac{d ^{2} z}{dx ^{2}} = - 2 < 0$

$∴ x = 12$ is a point of local maxima and z is maximum at $x = 12$ . $∴$ From equation (i), $y = 24 - 12 = 12$ Therefore, the two required numbers are 12 and 12.

Find two positive numbers x and y such that $x + y = 60$ and $xy^{3}$ is maximum.

Sol. The two numbers are x and y such that $x + y = 60$ . $∴ y = 60 - x$ Let $f (x) = xy^{3}$ . $\Rightarrow f (x) = x (60 - x)^{3}$ $∴ f^{'} (x) = (60 - x)^{3} - 3 x (60 - x)^{2}$ $= (60 - x)^{2} [60 - x - 3 x]$ $= (60 - x)^{2} (60 - 4 x)$ And

$f^{''} (x) = - 2 (60 - x) (60 - 4 x) - 4 (60 - x)^{2} = - 2 (60 - x) [60 - 4 x + 2 (60 - x)] = - 2 (60 - x) (180 - 6 x) = - 12 (60 - x) (30 - x)$

for maxima or minima Now $f^{'} (x) = 0 \Rightarrow x = 60$ or $x = 15$ When $x = 60$ ,

$f^{''} (x) = 0$

When $x = 15$ ,

$f^{''} (x) = - 12 (60 - 15) (30 - 15) = - 12 \times 45 \times 15 < 0 = - 8100 < 0$

∴ By second order derivative test, $x = 15$ is a point of local maxima of f(x). Thus, function $xy^{3}$ is maximum when $x = 15$ and $y = 60 - 15 = 45$ . Hence, the required numbers are 15 and 45.

Find two positive numbers x and y such that their sum is 35 and the product $x^{2} y^{5}$ is a maximum. Sol. Let one number be x. Then, the other number is $y = (35 - x)$ . Let $P (x) = x^{2} y^{5}$ . Then, we have

$P (x) = x^{2} (35 - x)^{5}$

$∴ P^{'} (x) = 2 x (35 - x)^{5} - 5 x^{2} (35 - x)^{4}$

$= x (35 - x)^{4} [2 (35 - x) - 5 x] = x (35 - x)^{4} (70 - 7 x) = 7 x (35 - x)^{4} (10 - x)$

And, P" (x) $= 7 (35 - x)^{4}$ $(10 - x) + 7 x [- (35 - x)^{4} - 4 (35 - x)^{3} (10 - x)]$ $= 7 (35 - x)^{4}$ $(10 - x) - 7 x (35 - x)^{4} - 28 x (35 - x)^{3} (10 - x)$ $= 7 (35 - x)^{3}$ $[(35 - x) (10 - x) - x (35 - x) - 4 x (10 - x)]$ $= 7 (35 - x)^{3}$ $[(350 - 45 x + x^{2} - 35 x + x^{2} - 40 x + 4 x^{2}]$ $= 7 (35 - x)^{3} (6 x^{2} - 120 x + 350)$ for maxima or minima Now, $P^{'} (x) = 0 \Rightarrow x = 0, 35, 10$ When $x = 35$ or $x = 0$ . This will make the product $x^{2} y^{5}$ equal to 0. $∴ x = 0$ and $y = 35$ cannot be the possible values of x. When $x = 10$ , we have

$P^{''} (x) = 7 (35 - 10)^{3} (6 \times 100 - 120 \times 10 + 350) = 7 (25)^{3} (- 250) < 0$

∴ By second order derivative test, $P (x)$ will be the maximum when $x = 10$ and $y = 35 - 10 = 25$ . Hence, the required numbers are 10 and 25.

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. Sol. Let one number be x. Then, the other number is $(16 - x)$ . Let the sum of the cubes of these numbers be denoted by S(x). Then, $S (x) = x^{3} + (16 - x)^{3}$ $∴ S^{'} (x) = 3 x^{2} - 3 (16 - x)^{2}, S^{''} (x)$

$= 6 x + 6 (16 - x)$

for maxima or minima $S^{'} (x) = 0$ $\Rightarrow 3 x^{2} - 3 (16 - x)^{2} = 0$ $\Rightarrow x^{2} - (16 - x)^{2} = 0$ $\Rightarrow x^{2} - 256 - x^{2} + 32 x = 0$ $\Rightarrow x = \frac{256}{32} = 8$

Now, $S^{'} (8) = 6 (8) + 6 (16 - 8)$

$= 48 + 48 = 96 > 0$

∴ By second order derivative test, $x = 8$ is the point of local minima of S. Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16 - 8 = 8$ .

A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Sol. Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18-2x) cm each and the height of the box is x cm.

Therefore, the volume $V (x)$ of the box is given by,

$V (x) = x (18 - 2 x)^{2}$

$∴ V^{'} (x) = (18 - 2 x)^{2} - 4 x (18 - 2 x)$

$= (18 - 2 x) [18 - 2 x - 4 x] = (18 - 2 x) (18 - 6 x) = 12 (9 - x) (3 - x)$

And

$V^{''} (x) = 12 [- (9 - x) - (3 - x)] = - 12 (9 - x + 3 - x) = - 12 (12 - 2 x) = - 24 (6 - x)$

for maxima or minima

$V^{'} (x) = 0 \Rightarrow x = 9 or x = 3$

If $x = 9$ , then the length and the breadth will become 0. $∴ x \neq = 9 \Rightarrow x = 3$ . Now, $V^{''} (3) = - 24 (6 - 3) = - 72 < 0$ ∴ By second order derivative test, $x = 3$ is the point of maxima of V.Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Sol. Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is $45 - 2 x$ , and the breadth is $24 - 2 x$ . Therefore, the volume V(x) of the box is given by,

$V (x) = x (45 - 2 x) (24 - 2 x) = x (1080 - 90 x - 48 x + 4 x^{2}) = 4 x^{3} - 138 x^{2} + 1080 x$

$∴ V^{'} (x) V^{''} (x) = 12 x^{2} - 276 x + 1080 = 12 (x^{2} - 23 x + 90) = 12 (x - 18) (x - 5) = 24 x - 276 = 12 (2 x - 23)$

for maxima or minima

$V^{'} (x) = 0 \Rightarrow x = 18 and x = 5$

It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. Thus, x cannot be equal to 18.

$∴ ∴ x \neq = 18 x = 5$

Now,

$V^{''} (5) = 12 (10 - 23) = 12 (- 13) = - 156 < 0$

∴ By second order derivative test, $x = 5$ is the point of maxima. Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. Sol. Let a rectangle of length $l$ and breadth b be inscribed in the given circle of radius a. Then, the diagonal passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have:

$(2 a)^{2} = ℓ^{2} + b^{2}$

$\Rightarrow b^{2} = 4 a^{2} - ℓ^{2}$ $\Rightarrow b = 4 a^{2} - ℓ^{2}$ ∴ Area of the rectangle,

$A = ℓ 4 a^{2} - ℓ^{2}$

$\Rightarrow A^{2} = ℓ^{2} (4 a^{2} - ℓ^{2}) = B ($ let $)$

If A is maximum B is also maximum

$∴ \frac{dB}{d ℓ} = 8 a^{2} ℓ - 4 ℓ^{3}$

$\Rightarrow \frac{d ^{2} B}{d ℓ ^{2}} = 8 a^{2} - 12 ℓ^{2}$ for maxima or minima $\frac{dB}{d ℓ} = 0$

$\Rightarrow \Rightarrow 8 a^{2} ℓ - 4 ℓ^{3} = 0 ℓ = 0 or ℓ = 2 a$

$\Rightarrow b = 4 a^{2} - 2 a^{2} = 2 a^{2} = 2 a$

Now, when $ℓ = 2 a$

$\frac{d ^{2} B}{d ℓ ^{2}} = 8 a^{2} - 12 (2 a^{2}) = - 16 a^{2} < 0$

∴ By the second order derivative test, when $ℓ = 2 a$ , then the area of the rectangle is the maximum. Since $ℓ = b = 2 a$ , the rectangle is a square. Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base. Sol. Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by,

$S = 2 π r^{2} + 2 π r$

$\Rightarrow h = \frac{S - 2 π r ^{2}}{2 π r}$ $\Rightarrow h = \frac{S}{2 π r} - r$

Let V be the volume of the cylinder, Then,

$V = π r^{2} h = π r^{2} [\frac{S}{2 π r} - r] = \frac{Sr}{2} - π r^{3}$

[By eq.(1)] Then, $\frac{dV}{dr} = \frac{S}{2} - 3 π r^{2}, \frac{d ^{2} V}{dr ^{2}} = - 6 π r$ for maxima or minima,

$\frac{dV}{dr} = 0 \Rightarrow \frac{S}{2} = 3 π r^{2} \Rightarrow r^{2} = \frac{S}{6 π}$

When $r^{2} = \frac{S}{6 π}$ , then $\frac{d ^{2} V}{dr ^{2}} = - 6 π (\frac{S}{6 π}) < 0$

∴ By second order derivative test, the volume is the maximum when $r^{2} = \frac{S}{6 π}$ Now, when $r^{2} = \frac{S}{6 π}$ , or $S = 6 π r^{2}$ then $h = \frac{6 π r ^{2}}{2 π r} - r = 3 r - r = 2 r$ Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area? Sol. Let r and h be the radius and height of the cylinder respectively, then volume (V) of the cylinder is given by,

$V = π r^{2} h = 100 (given)$

$∴ h = \frac{100}{π r ^{2}}$

Surface area (S) of the cylinder is given by,

$S = 2 π r^{2} + 2 π rh = 2 π r^{2} + \frac{200}{r}$

$∴ \frac{dS}{dr} = 4 π r - \frac{200}{r ^{2}}, \frac{d ^{2} S}{dr ^{2}} = 4 π + \frac{400}{r ^{3}}$ for maxima or minima, $\frac{dS}{dr} = 0 \Rightarrow 4 π r = \frac{200}{r ^{2}}$ $\Rightarrow r^{3} = \frac{200}{4 π} = \frac{50}{π} \Rightarrow r^{3} = \frac{50}{π} \Rightarrow r = (\frac{50}{π})^{\frac{1}{3}}$

Now, it is observed that when

$r = (\frac{50}{π})^{\frac{1}{3}}, \frac{d ^{2} S}{dr ^{2}} > 0$

By second order derivative test, the surface area is the minimum when the radius of the cylinder is $(\frac{50}{π})^{\frac{1}{3}} cm$ Now, $\frac{d ^{2} S}{dr ^{2}} = \frac{400}{r ^{3}} + 4 π = 8 π + 4 π = 12 π > 0$ When $r = (\frac{50}{π})^{\frac{1}{3}}$

$h = \frac{100}{π ( \frac{50}{π} ) ^{\frac{2}{3}}} = \frac{2 \times 50}{( 50 ) ^{\frac{2}{3}} ( \frac{1}{π} ) ^{\frac{2}{3}} π} = 2 (\frac{50}{π})^{\frac{1}{3}} cm .$

Hence, the required dimensions of the can which has the minimum surface area is given

$by radius = (\frac{50}{π})^{\frac{1}{3}} cm$

and height $= 2 (\frac{50}{π})^{\frac{1}{3}} cm$ .

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? Sol. Let a piece of length $ℓ$ be cut from the given wire to make a square. Then, the other piece of wire to be made into a circle is of length $(28 - ℓ) m$ . Now, side of square $= \frac{ℓ}{4}$ . Let r be the radius of the circle. Then, $2 π r = 28 - ℓ \Rightarrow r = \frac{1}{2 π} (28 - ℓ)$ The combined areas of the square and the circle (A) is given by, $A = (side of the square)^{2} + π r^{2}$

$= \frac{ℓ ^{2}}{16} + π [\frac{1}{2 π} (28 - ℓ)]^{2} = \frac{ℓ ^{2}}{16} + \frac{1}{4 π} (28 - ℓ)^{2}$

$∴ \frac{dA}{d ℓ} = \frac{2 ℓ}{16} + \frac{2}{4 π} (28 - ℓ) (- 1) = \frac{ℓ}{8} - \frac{1}{2 π} (28 - ℓ)$ $\Rightarrow \frac{dA}{d ℓ} = \frac{ℓ}{8} - \frac{1}{2 π} (28 - ℓ)$ $\Rightarrow \frac{d ^{2} A}{d ℓ ^{2}} = \frac{1}{8} + \frac{1}{2 π}$

for maxima or minima

$\frac{dA}{d ℓ} = 0 \Rightarrow \frac{ℓ}{8} - \frac{1}{2 π} (28 - ℓ) = 0$

$\Rightarrow \frac{π ℓ - 4 ( 28 - ℓ )}{8 π} = 0 \Rightarrow (π + 4) ℓ - 112 = 0$ $\Rightarrow ℓ = \frac{112}{π + 4}$ Thus, when $ℓ = \frac{112}{π + 4}, \frac{d ^{2} A}{d ℓ ^{2}} > 0$

∴ By second order derivative test, the area (A) is the minimum when $ℓ = \frac{112}{π + 4}$ . Hence, the combined area is the minimum when the length of the wire in making the square is $\frac{112}{π + 4} m$ while the length of the wire in making the circle is $28 - \frac{112}{π + 4} = \frac{28 π}{π + 4} m$

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere. Sol. Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R. Let V be the volume of the cone.

Then, $V = \frac{1}{3} π r^{2} h$ Height of the cone is given by,

$h = R + AB = R + R^{2} - r^{2}$

[ABC is a right triangle]

$\Rightarrow r^{2} = 2 h R - h^{2}$

$∴ \Rightarrow ∴ V = \frac{1}{3} π h (2 hR - h^{2}) V = \frac{1}{3} π (2 h^{2} R - h^{3}) \frac{dV}{dh} = \frac{1}{3} π (4 hR - 3 h^{2}) and \frac{d ^{2} V}{dh ^{2}} = \frac{1}{3} π (4 R - 6 h)$

for maxima or minima

$\Rightarrow \Rightarrow ∴ \frac{d V}{d h} = 0 4 h R - 3 h^{2} = 0 h = 0 or h = \frac{4 R}{3} rejecting h = 0 h = \frac{4 R}{3} When h = \frac{4 R}{3} then \frac{d ^{2} V}{d h ^{2}} = \frac{1}{3} π (4 R - 8 R) = - \frac{4 π R}{3} < 0 from (1) r^{2} = 2 h R - h^{2} when h = \frac{4 R}{3}, r^{2} = \frac{8 R ^{2}}{9}$

By second order derivative test, the volume of the cone is the maximum when $r^{2} = \frac{8}{9} R^{2}$ . Therefore, volume of cone

$V = \frac{1}{3} π (\frac{8}{9} R^{2}) (\frac{4}{3} R) = \frac{8}{27} (\frac{4}{3} π R^{3}) = \frac{8}{27} \times (Volume of the sphere)$

Hence, the volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$ the volume of the sphere.

Show that the right circular cone of least curved surface and given volume has an altitude equal to $2$ times the radius of the base. Sol. Let r and h be the radius and the height (altitude) of the cone respectively. Then, the volume (V) of the cone is given as :

$V = \frac{1}{3} π r^{2} h \Rightarrow h = \frac{3 V}{π r ^{2}}$

The surface area (S) of the cone is given by, $S = π r ℓ$ (where $ℓ$ is the slant height) $\Rightarrow S^{2} = π r^{2} ℓ^{2}$

$\Rightarrow S^{2} S^{2} = π^{2} r^{2} (r^{2} + h^{2}) (∵ ℓ^{2} = r^{2} + h^{2}) = π^{2} r^{2} (r^{2} + \frac{9 v ^{2}}{π ^{2} r ^{4}}) = π^{2} r^{4} + \frac{9 V ^{2}}{r ^{2}} = z (let)$

If S is minimum $S^{2} (z)$ is also minimum $∴ \frac{dz}{dr} = 4 π^{2} r^{3} - \frac{18 v ^{2}}{r ^{3}}$ and $\frac{d ^{2} z}{dr ^{2}} = 12 π^{2} r^{2} + \frac{54 v ^{2}}{r ^{4}}$ for maxima or minima $\frac{dz}{dr} = 0$ $\Rightarrow 4 π^{2} r^{3} - \frac{18 v ^{2}}{r ^{3}} = 0$

$\Rightarrow r^{6} At \frac{d ^{2} z}{d r ^{2}} = (\frac{9 v ^{2}}{2 π ^{2}}) 9 V^{2} = 2 π^{2} r^{6}, = \frac{54}{r ^{4}} (\frac{2 π ^{2} r ^{6}}{9}) + 12 π^{2} r^{2} = 12 π^{2} r^{2} + 12 π^{2} r^{2} = 24 π^{2} r^{2} > 0$

By second order derivative test, the surface area of the cone is the least when $r^{6} = \frac{9 V ^{2}}{2 π ^{2}}$ . When $r^{6} = \frac{9 V ^{2}}{2 π ^{2}}$ , $h = \frac{3 V}{π r ^{2}} = \frac{3}{π r ^{2}} (\frac{2 π ^{2} r ^{6}}{9})^{\frac{1}{2}} = \frac{3}{π r ^{2}} \cdot \frac{2 π r ^{3}}{3} = 2 r$ . Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to $2$ times the radius of the base.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $tan^{- 1} 2$ . Sol. Let $θ$ be the semi-vertical angle of the cone. It is clear that $θ \in [0, \frac{π}{2}]$ .

Let $r, h$ and $ℓ$ be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant. Now, $r = l sin θ$ and $h = l cos θ$ The volume (V) of the cone is given by,

$V ∴ \frac{dV}{d θ} = \frac{1}{3} π r^{2} h = \frac{1}{3} π (ℓ^{2} sin^{2} θ) (ℓ cos θ) = \frac{1}{3} π ℓ^{3} sin^{2} θ cos θ = \frac{ℓ ^{3} π}{3} [sin^{2} θ (- sin θ) + cos θ (2 sin θ cos θ)] = \frac{ℓ ^{3} π}{3} [- sin^{3} θ + 2 sin θ cos^{2} θ]$

$\frac{d ^{2} V}{d θ ^{2}} = \frac{ℓ ^{3} π}{3} [- 3 sin^{2} θ cos θ + 2 cos^{3} θ - 4 sin^{2} θ cos θ] = \frac{ℓ ^{3} π}{3} [2 cos^{3} θ - 7 sin^{2} θ cos θ]$

for maxima or minima $\frac{dV}{d θ} = 0$ $\Rightarrow sin^{3} θ = 2 sin θ cos^{2} θ$ $\Rightarrow tan^{2} θ = 2 \Rightarrow tan θ = 2$ $\Rightarrow θ = tan^{- 1} 2$

Now, when $θ = tan^{- 1} 2$ , then $tan^{2} θ = 2$ or $sin^{2} θ = 2 cos^{2} θ$ . Then, we have :

$\frac{d ^{2} V}{d θ ^{2}} = \frac{ℓ ^{3} π}{3} [2 cos^{3} θ - 14 cos^{3} θ] = - 4 π ℓ^{3} cos^{3} θ < 0$

for $θ \in [0, \frac{π}{2}]$ By second derivative test, the volume (V) is the maximum when $θ = tan^{- 1} 2$ Hence, for a given slant height , the semivertical angle of the cone of the maximum volume is $tan^{- 1} 2$ .

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $sin^{- 1} (\frac{1}{3})$ Sol. $s = π r^{2} + π r ℓ$ (given)

$\Rightarrow ℓ = \frac{s - π r ^{2}}{π r}$

Let v be the volume $v = \frac{1}{3} π r^{2} h$

$v^{2} = \frac{1}{9} π^{2} r^{4} h^{2} = \frac{1}{9} π^{2} r^{4} (ℓ^{2} - r^{2}) [∵ h^{2} = ℓ^{2} - r^{2}] = \frac{1}{9} π^{2} r^{4} [(\frac{s - π r ^{2}}{π r})^{2} - r^{2}] = \frac{1}{9} π^{2} r^{4} [\frac{( s - π r ^{2} ) ^{2}}{π ^{2} r ^{2}} - \frac{r ^{2}}{1}] [From (1)] = \frac{1}{9} r^{2} [(s - π r^{2})^{2} - π^{2} r^{4}] = \frac{1}{9} r^{2} [s^{2} + π^{2} r^{4} - 2 s π r^{2} - π^{2} r^{4}] = \frac{1}{9} r^{2} [s^{2} - 2 s π r^{2}]$

$\Rightarrow z = \frac{1}{9} [s^{2} r^{2} - 2 s π r^{4}] [∵ v^{2} = z]$ Now, $\frac{dz}{dr} = \frac{1}{9} [2 rs^{2} - 8 s π r^{3}]$

$\frac{dz}{dr} = 0 [For maxima or Minima] 0 = \frac{1}{9} [2 rs^{2} - 8 s π r^{3}] 8 s π r^{3} = 2 rs^{2} \Rightarrow 4 π r^{2} = s Now, \frac{d ^{2} z}{dr ^{2}} = \frac{1}{9} [2 s^{2} - 24 s π r^{2}] \frac{d ^{2} z}{dr ^{2}}]_{r^{2} = \frac{s}{4 π}} = \frac{1}{9} [2 s^{2} - 24 s π \cdot \frac{s}{4 π}] = - \frac{4}{9} s^{2} < 0$

∴ By second derivative test, volume v is maximum at $s = 4 π r^{2}$ Now, we have $s = π r ℓ + π r^{2}$

$4 π r^{2} = π r ℓ + π r^{2}$

$\Rightarrow 3 π r^{2} = π r ℓ \Rightarrow 3 r = ℓ \Rightarrow \frac{r}{ℓ} = \frac{1}{3} \Rightarrow sin α = \frac{1}{3}$

$∴ α = sin^{- 1} (\frac{1}{3})$

Choose the correct answer in the following questions from 27 to 29.

The point on the curve $x^{2} = 2 y$ which is nearest of the point $(0, 5)$ is? (A) $(22, 4)$ (B) $(22, 0)$ (C) $(0, 0)$ (D) $(2, 2)$ Sol. (A) Given curve is $x^{2} = 2 y$ Let the point ( $x, y$ ) on the curve nearest to the point (0, 5). Now distance (D) between (x, y) and $(0, 5)$ is

$D = x^{2} + (y - 5)^{2}$

$\Rightarrow D = 2 y + (y - 5)^{2} (∵ x^{2} = 2 y)$

$D^{2} = 2 y + y^{2} + 25 - 10 y$

$\Rightarrow D^{2} = y^{2} - 8 y + 25$ Let $D^{2} = z$ Then $z = y^{2} - 8 y + 25$

$z^{'} = 2 y - 8 and z^{''} = 2$

for maxima or minima Now, $z^{'} = 0 \Rightarrow y = 4$ At $y = 4, z^{''} = 2 > 0$ i.e. $y = 4$ is point of minima. At $y = 4, x = \pm 22$ Hence, required point is $(\pm 22, 4)$ . The correct answer is A.

For all real values of x, the minimum value of $\frac{1 - x + x ^{2}}{1 + x + x ^{2}}$ is ? (A) 0 (B) 1 (C) 3 (D) $\frac{1}{3}$ Sol. (D) Let $f (x) = \frac{1 - x + x ^{2}}{1 + x + x ^{2}}$ .

$∴ f^{'} (x) = \frac{( 1 + x + x ^{2} ) ( - 1 + 2 x ) - ( 1 - x + x ^{2} ) ( 1 + 2 x )}{( 1 + x + x ^{2} ) ^{2}} = \frac{- 1 + 2 x - x + 2 x ^{2} - x ^{2} + 2 x ^{3} - 1 - 2 x + x + 2 x ^{2} - x ^{2} - 2 x ^{3}}{( 1 + x + x ^{2} ) ^{2}} = \frac{2 x ^{2} - 2}{( 1 + x + x ^{2} ) ^{2}} = \frac{2 ( x ^{2} - 1 )}{( 1 + x + x ^{2} ) ^{2}} for maxima or minima f^{'} (x) = 0 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$

Now,

$f^{''} (x) = \frac{2 [ ( 1 + x + x ^{2} ) ^{2} ( 2 x ) - ( x ^{2} - 1 ) ( 2 ) ( 1 + x + x ^{2} ) ( 1 + 2 x ) ]}{( 1 + x + x ^{2} ) ^{4}} = \frac{4 ( 1 + x + x ^{2} ) [ ( 1 + x + x ^{2} ) x - ( x ^{2} - 1 ) ( 1 + 2 x ) ]}{( 1 + x + x ^{2} ) ^{4}} = \frac{4 [ x + x ^{2} + x ^{3} - x ^{2} - 2 x ^{3} + 1 + 2 x ]}{( 1 + x + x ^{2} ) ^{3}} = \frac{4 [ 1 + 3 x - x ^{3} ]}{( 1 + x + x ^{2} ) ^{3}} And, f^{''} (1) = \frac{4 [ 1 + 3 - 1 ]}{( 1 + 1 + 1 ) ^{3}} = \frac{4 ( 3 )}{( 3 ) ^{3}} = \frac{4}{9} > 0 Also, f^{''} (- 1) = \frac{4 [ 1 - 3 + 1 ]}{( 1 - 1 + 1 ) ^{3}} = 4 (- 1) = - 4 < 0$

By second order derivative test, f is minimum at $x = 1$ and the minimum value is given by

$f (1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3} . The correct answer is D .$

The maximum value of

$[x (x - 1) + 1]^{\frac{1}{3}}, 0 \leq x \leq 1 is?$

(A) $(\frac{1}{3})^{\frac{1}{3}}$ (B) $\frac{1}{2}$ (C) 1 (D) 0 Sol. (C) Let $f (x) = [x (x - 1) + 1]^{\frac{1}{3}}$

$∴ f^{'} (x) = \frac{2 x - 1}{3 [ x ( x - 1 ) + 1 ] ^{\frac{2}{3}}}$

for maxima or minima

$f^{'} (x) = 0 \Rightarrow x = \frac{1}{2}$

Then, we evaluate the value of f at critical point $x = \frac{1}{2}$ and at the end points of the interval [0, 1] {i.e., at $x = 0$ and $x = 1$ }.

$f (0) = [0 (0 - 1) + 1]^{\frac{1}{3}} = 1 f (1) = [1 (1 - 1) + 1]^{\frac{1}{3}} = 1 f (\frac{1}{2}) = [\frac{1}{2} (- \frac{1}{2}) + 1]^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}}$

Hence, we can conclude that that maximum value of f in the interval $[0, 1]$ is 1 . The correct answer is C.

MISCELLANEOUS EXERCISE

Show that the function given by $f (x) = \frac{l o g x}{x}$ has maximum at $x = e$ . Sol. The given function is $f (x) = \frac{l o g x}{x}$ .

$f^{'} (x) = \frac{x ( \frac{1}{x} ) - l o g x}{x ^{2}} = \frac{1 - l o g x}{x ^{2}}$

for maxima or minima $f^{'} (x) = 0$

$\Rightarrow \frac{1 - l o g x}{x ^{2}} = 0 \Rightarrow lo g x = 1$

$\Rightarrow lo g x = lo g e \Rightarrow x = e$ Now, $f^{''} (x) = \frac{x ^{2} ( - \frac{1}{x} ) - ( 1 - l o g x ) ( 2 x )}{x ^{4}}$

$= \frac{- x - 2 x ( 1 - l o g x )}{x ^{4}} = \frac{- 3 + 2 l o g x}{x ^{3}}$

at $x = e, f^{''} (e) = \frac{- 3 + 2 l o g e}{e ^{3}} = \frac{- 3 + 2}{e ^{3}} = \frac{- 1}{e ^{3}} < 0$ Therefore, by second order derivative test, f(x) is maximum at $x = e$ .

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ? Sol. Let $△ ABC$ be isosceles where BC is the base of fixed length b . Let the length of the two equal sides of $△ ABC$ be a. Draw AD ⟂ BC

Now, in $△ ADC$ , by applying the Pythagoras theorem, we have : $AD = a^{2} - \frac{b ^{2}}{4}$ Area of triangle ABC is $A = \frac{1}{2} BC \cdot AD = \frac{b}{2} a^{2} - \frac{b ^{2}}{4}$ The rate of change of the area with respect of time (t) is given by, $\frac{dA}{dt} = \frac{1}{2} b . \frac{2 a}{2 a ^{2} - \frac{b ^{2}}{4}} \frac{da}{dt} = \frac{ab}{4 a ^{2} - b ^{2}} \frac{da}{dt}$ It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. $\Rightarrow \frac{da}{dt} = - 3 cm / s$ $∴ \frac{dA}{dt} = \frac{- 3 ab}{4 a ^{2} - b ^{2}}$ Then, when $a = b$ , we have :

$\frac{dA}{dt} = \frac{- 3 b ^{2}}{4 b ^{2} - b ^{2}} = \frac{- 3 b ^{2}}{3 b ^{2}} = - 3 b cm^{2} / s$

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $3 b cm^{2} / s$ .

Find the intervals in which the function f given by $f (x) = \frac{4 s i n x - 2 x - x c o s x}{2 + c o s x}$ is (i) increasing (ii) decreasing Sol. $f (x) = \frac{4 s i n x - 2 x - x c o s x}{2 + c o s x}$ $f^{'} (x) = \frac{( 2 + c o s x ) ( 4 c o s x - 2 - c o s x + x s i n x ) - ( 4 s i n x - 2 x - x c o s x ) ( - s i n x )}{( 2 + c o s x ) ^{2}}$

PRACTICAL BOOKS for BRINGING LEARNING & LIFE

SHOP NOW $= \frac{( 2 + c o s x ) ( 3 c o s x - 2 + x s i n x ) + s i n x ( 4 s i n x - 2 x - x c o s x )}{( 2 + c o s x ) ^{2}}$ $= \frac{6 c o s x - 4 + 2 x s i n x + 3 c o s ^{2} x - 2 c o s x + x s i n x c o s x + 4 s i n ^{2} x - 2 x s i n x - x s i n x c o s x}{( 2 + c o s x ) ^{2}}$ $= \frac{4 c o s x - 4 + 3 c o s ^{2} x + 4 s i n ^{2} x}{( 2 + c o s x ) ^{2}}$ $= \frac{4 c o s x - 4 + 3 c o s ^{2} x + 4 - 4 c o s ^{2} x}{( 2 + c o s x ) ^{2}}$ $= \frac{4 c o s x - c o s ^{2} x}{( 2 + c o s x ) ^{2}} = \frac{c o s x ( 4 - c o s x )}{( 2 + c o s x ) ^{2}}$ Now, $f^{'} (x) = 0 \Rightarrow cos x = 0$ or $cos x = 4$ But, $cos x \neq = 4 \Rightarrow x = \frac{π}{2}, \frac{3 π}{2}$ Now, $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ divides $[0, 2 π]$ into three intervals i.e.,

$[0, \frac{π}{2}], [\frac{π}{2}, \frac{3 π}{2}] and [\frac{3 π}{2}, 2 π] .$

In intervals $[0, \frac{π}{2}]$ and $[\frac{3 π}{2}, 2 π], f^{'} (x) > 0$ Thus, $f (x)$ is increasing for $[0, \frac{π}{2}]$ and $[\frac{3 π}{2}, 2 π]$ In the interval $[\frac{π}{2}, \frac{3 π}{2}], f^{'} (x) \leq 0$ . Thus, $f (x)$ is decreasing for $[\frac{π}{2}, \frac{3 π}{2}]$ .

Find the intervals in which the function f given by $f (x) = x^{3} + \frac{1}{x ^{3}}, x \neq = 0$ is

(i) increasing (ii) decreasing

Sol. $f (x) = x^{3} + \frac{1}{x ^{3}}$

$∴ f^{'} (x) = 3 x^{2} - \frac{3}{x ^{4}} = \frac{3 x ^{6} - 3}{x ^{4}}$

Then, $f^{'} (x) = 0$ '

$\Rightarrow 3 x^{6} - 3 = 0 \Rightarrow x^{6} = 1 \Rightarrow x = \pm 1$

Now, the points $x = 1$ and $x = - 1$ divide the real line into three intervals

i.e., $(- \infty, - 1], [- 1, 1]$ and $[1, \infty)$ .

In intervals $(- \infty, - 1]$ and $[1, \infty) f^{'} (x) \geq 0$ $∴ f (x)$ is increasing in $(- \infty, - 1] \cup [1, \infty)$ In interval $[- 1, 1] f^{'} (x) < 0$ . $∴ f (x)$ is decreasing in $[- 1, 1]$

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with its vertex at one end of the major axis. Sol. The given ellipse is $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , Its parametric form is $x = a cos θ, y = b sin θ$ Let the major axis be along the x -axis. Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0) Since the ellipse is symmetrical with respect to the x-axis and y-axis, we can assume the coordinates of A to be ( $- a cos θ, b sin θ$ ) and the coordinates of B to be ( $- a cos θ, - b sin θ$ )

$\Rightarrow AB = 2 b sin θ$ and $DC = (a + a cos θ)$ the area of triangle ABC is given by,

$A = \frac{1}{2} AB \cdot DC$

$\Rightarrow A = \frac{1}{2} (2 b sin θ) \cdot (a + a cos θ)$ $\Rightarrow A = ab sin θ (1 + cos θ)$ $\Rightarrow \frac{dA}{d θ} = ab [sin θ (- sin θ) + (1 + cos θ) cos θ]$

$\Rightarrow \frac{dA}{d θ} = ab [cos θ + cos^{2} θ - sin^{2} θ]$ and $\frac{d ^{2} A}{d θ ^{2}} = ab [- sin θ - 2 sin θ cos θ - 2 sin θ cos θ]$ $\Rightarrow \frac{d ^{2} A}{d θ ^{2}} = - ab [sin θ + 4 sin θ cos θ]$ for maxima or minima $\frac{dA}{d θ} = 0$ $\Rightarrow ab [cos θ + cos^{2} θ - sin^{2} θ] = 0$ $\Rightarrow cos θ + cos^{2} θ - 1 + cos^{2} θ = 0$ $\Rightarrow 2 cos^{2} θ + cos θ - 1 = 0$ $\Rightarrow cos θ = - 1$ or $cos θ = \frac{1}{2} \Rightarrow θ = π$ or $θ = \frac{π}{3}$ rejecting $θ = π \Rightarrow θ = \frac{π}{3}$ Now at $θ = \frac{π}{3}$ ,

$\frac{d ^{2} A}{d θ ^{2}} = - ab [sin \frac{π}{3} + 4 sin \frac{π}{3} cos \frac{π}{3}] = - ab [\frac{3}{2} + 4 \cdot \frac{3}{2} \cdot \frac{1}{2}] < 0$

∴ By second derivative test, Area (A) of isosceles triangle is maximum when $θ = \frac{π}{3}$ Maximum area of the triangle is given by, $A = ab sin θ (1 + cos θ)$

$= absin \frac{π}{3} (1 + cos \frac{π}{3})$

A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is $8 m^{3}$ . If building of tank costs ₹70 per square meters for the base and ₹45 per square metre for sides. What is the cost of least expensive tank? Sol. Let $ℓ, b$ and h represent the length, breadth, and height of the tank respectively. Then, we have height $(h) = 2 m$ Volume of the tank $= 8 m^{3}$ ; Volume of the tank $= ℓ \times b \times h$ $\Rightarrow 8 = ℓ \times b \times 2 \Rightarrow ℓ b = 4 \Rightarrow b = \frac{4}{ℓ}$

Now, area of the base $= ℓ b = 4$ Area of the 4 walls $(A) = 2 h (ℓ + b)$

$∴ A = 4 (ℓ + \frac{4}{ℓ}) \Rightarrow \frac{dA}{d ℓ} = 4 (1 - \frac{4}{ℓ ^{2}})$ for maxima or minima $\frac{dA}{d ℓ} = 0 \Rightarrow 1 - \frac{4}{ℓ ^{2}} = 0 \Rightarrow ℓ^{2} = 4 \Rightarrow ℓ = \pm 2$ However, the length cannot be negative. Therefore, we have $ℓ = 2$ $∴ b = \frac{4}{ℓ} = \frac{4}{2} = 2$ Now, $\frac{d ^{2} A}{d ℓ ^{2}} = \frac{32}{ℓ ^{3}}$ When $ℓ = 2, \frac{d ^{2} A}{d ℓ ^{2}} = \frac{32}{8} = 4 > 0$ . Thus, by second order derivative test, the area is the minimum when $ℓ = 2$ . We have $ℓ = b = h = 2$ . ⇒ Cost of building the base $= ₹70 \times (ℓ b)$ = ₹70 (4) = ₹280 Cost of building the walls $= ₹2 h (ℓ + b) \times 45$

$= ₹90 (2) (2 + 2) = ₹720$

Required total cost $= ₹ (280 + 720) = ₹1000$ Hence, the total cost of the tank will be ₹1000.

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle. Sol. Let 'r' be the radius of the circle and 'a' be the side of the square. Then, we have $2 π r + 4 a = k$ (where k is constant) $\Rightarrow a = \frac{k - 2 π r}{4}$

The sum of the areas of the circle and the square (A) is given by,

$A = π r^{2} + a^{2} = π r^{2} + \frac{( k - 2 π r ) ^{2}}{16}$

$\Rightarrow \frac{dA}{dr} = 2 π r + \frac{2 ( k - 2 π r ) ( - 2 π )}{16}$

$= 2 π r - \frac{π ( k - 2 π r )}{4}$

for maxima or minima

$\frac{dA}{dr} = 0$

$\Rightarrow 2 π r = \frac{π ( k - 2 π r )}{4}$

$8 r = k - 2 π r \Rightarrow (8 + 2 π) r = k$

$\Rightarrow r = \frac{k}{8 + 2 π} = \frac{k}{2 ( 4 + π )}$

Now, $\frac{d ^{2} A}{dr ^{2}} = 2 π + \frac{π ^{2}}{2}$ When $r = \frac{k}{2 ( 4 + π )}, \frac{d ^{2} A}{dr ^{2}} > 0$ .

The sum of the areas is least when

$r = \frac{k}{2 ( 4 + π )} .$

When $r = \frac{k}{2 ( 4 + π )}$

$a = \frac{k - 2 π [ \frac{k}{2 ( 4 + π )} ]}{4} = \frac{k - \frac{π k}{( 4 + π )}}{4} = \frac{4 k + π k - π k}{4 ( 4 + π )} = \frac{k}{4 + π} = 2 r$

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m . Find the dimensions of the window to admit maximum light through the whole opening. Sol. Let 2x and y be the length and breadth of the rectangular window respectively.

Radius of the semicircular opening $= x$ It is given that the perimeter of the window is 10 m. $∴ 2 x + 2 y + π x = 10 \Rightarrow y = 5 - x - \frac{π x}{2}$

Now, Area of the window is given by,

$A = 2 x y + \frac{π x ^{2}}{2}$

$\Rightarrow A = 2 x (5 - x - \frac{π x}{2}) + \frac{π x ^{2}}{2}$ $\Rightarrow A = 10 x - 2 x^{2} - π x^{2} + \frac{π x ^{2}}{2}$ $\Rightarrow A = 10 x - 2 x^{2} - \frac{π x ^{2}}{2}$ $∴ \frac{dA}{dx} = 10 - 4 x - π x$ $∴ \frac{d ^{2} A}{dx ^{2}} = - 4 - π$ for maxima or minima

$\frac{dA}{dx} = 0 \Rightarrow 10 - 4 x - π x = 0 \Rightarrow x = \frac{10}{π + 4}$

Thus, when $x = \frac{10}{π + 4}$ , then $\frac{d ^{2} A}{dx ^{2}} < 0$ . Therefore, by second order derivative test, the area is the maximum when length $2 x = \frac{20}{π + 4} m$

Now, $y = 5 - \frac{10}{π + 4} - \frac{π}{2} (\frac{10}{π + 4}) \Rightarrow y = \frac{10}{π + 4}$ Hence, the required dimensions of the window to admit maximum light is given by length $= \frac{20}{π + 4} m$ and breadth $= \frac{10}{π + 4} m$ .

A point on the hypotenuse of a triangle is at distance 'a' and 'b' from the sides of the triangle. Show that the minimum length of the hypotenuse is $(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$ Sol. Let $△ ABC$ be right-angled at B . Let $AB = x$ and $BC = y$ .

Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively. Let $∠ C = θ$ . We have, $AC = x^{2} + y^{2}$ Now,

$PC = b cosec θ and, A P = a sec θ A C = A P + PC A C = b cosec θ + a sec θ$

and

$\frac{d ^{2} ( AC )}{d θ ^{2}} = b (cosec θ cot^{2} θ + cosec^{3} θ) + a (sec^{3} θ + sec θ tan^{2} θ)$

for maxima or minima

$\frac{d ( AC )}{d θ} = 0 \Rightarrow a sec θ tan θ = b cosec θ cot θ$

$\Rightarrow \frac{a}{cos θ} \cdot \frac{sin θ}{cos θ} = \frac{b}{sin θ} \frac{cos θ}{sin θ} \Rightarrow a sin^{3} θ = b cos^{3} θ \Rightarrow (a)^{\frac{1}{3}} sin θ = (b)^{\frac{1}{3}} cos θ \Rightarrow tan θ = (\frac{b}{a})^{\frac{1}{3}} ∴ sin θ = \frac{( b ) ^{\frac{1}{3}}}{a ^{\frac{2}{3}} + b ^{\frac{2}{3}}} and cos θ = \frac{( a ) ^{\frac{1}{3}}}{a ^{\frac{2}{3}} + b ^{\frac{2}{3}}}$

Since, $0 < θ < \frac{π}{2}$ , so trigonometric ratios are positive. Also, $a > 0$ and $b > 0 \frac{d ^{2} ( AC )}{d θ ^{2}}$ is positive. It can be clearly shown that $\frac{d ^{2} ( AC )}{d θ ^{2}} > 0$ when

$tan θ = (\frac{b}{a})^{\frac{1}{3}} .$

Therefore, by second order derivative test, the length of the hypotenuse is minimum when

$tan θ = (\frac{b}{a})^{\frac{1}{3}} .$

Now, when $tan θ = (\frac{b}{a})^{\frac{1}{3}}$ , we have :

$A C = \frac{b a ^{\frac{2}{3}} + b ^{\frac{2}{3}}}{b ^{\frac{1}{3}}} + \frac{a a ^{\frac{2}{3}} + b ^{\frac{2}{3}}}{a ^{\frac{1}{3}}}$

[Using (1) and (2)]

$= a^{\frac{2}{3}} + b^{\frac{2}{3}} (b^{\frac{2}{3}} + a^{\frac{2}{3}}) = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$

Hence, the minimum length of the hypotenuse is $(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$ .

Find the points at which the function f given by $f (x) = (x - 2)^{4} (x + 1)^{3}$ has (i) local maxima (ii) local minima (iii) point of inflexion

Sol. The given function is $f (x) = (x - 2)^{4} (x + 1)^{3}$ .

$∴ f^{'} (x) = 4 (x - 2)^{3} (x + 1)^{3} + 3 (x + 1)^{2} (x - 2)^{4}$ $= (x - 2)^{3} (x + 1)^{2} [4 (x + 1) + 3 (x - 2)]$ $= (x - 2)^{3} (x + 1)^{2} (7 x - 2)$ For maxima or minima $f^{'} (x) = 0 \Rightarrow (x - 2)^{3} (x + 1)^{2} (7 x - 2) = 0$ $f^{'} (x) = 0 \Rightarrow x = - 1$ and $x = \frac{2}{7}$ or $x = 2$ Now, for values of x close to $\frac{2}{7}$ and to the left of $\frac{2}{7}, f^{'} (x) > 0$ . Also, for values of x close to $\frac{2}{7}$ and to the right of $\frac{2}{7}, f^{'} (x) < 0$ . Thus, $x = \frac{2}{7}$ is the point of local maxima. Now, for values of x close to 2 and to the left of 2, $f^{'} (x) < 0$ . Also, for values of x close to 2 and to the right of $2, f^{'} (x) > 0$ Thus, $x = 2$ is the point of local minima. Now, as the value of x varies through $- 1, f^{'} (x)$ does not changes its sign. Thus, $x = - 1$ is the point of inflexion.

Find the absolute maximum and minimum values of the function f given by

$f (x) = cos^{2} x + sin x, x \in [0, π]$

Sol. $f (x) = cos^{2} x + sin x$

$f^{'} (x) = 2 cos x (- sin x) + cos x = - 2 sin x cos x + cos x$

Now, $f^{'} (x) = 0$ $\Rightarrow 2 sin x cos x = cos x$ $\Rightarrow cos x (2 sin x - 1) = 0$ $\Rightarrow sin x = \frac{1}{2}$ or $cos x = 0 \Rightarrow x = \frac{π}{6}$ or $\frac{π}{2}$ as $x \in [0, π]$ Now, evaluating the value of f at critical points $x = \frac{π}{2}$ and $x = \frac{π}{6}$ and the end points of the interval $[0, π]$ (i.e., at $x = 0$ and $x = π$ ), we have: $\Rightarrow f (\frac{π}{6}) = cos^{2} \frac{π}{6} + sin \frac{π}{6} = (\frac{3}{2})^{2} + \frac{1}{2} = \frac{5}{4}$ $\Rightarrow f (0) = cos^{2} 0 + sin 0 = 1 + 0 = 1$ $\Rightarrow f (π) = cos^{2} π + sin π = (- 1)^{2} + 0 = 1$ $\Rightarrow f (\frac{π}{2}) = cos^{2} \frac{π}{2} + sin \frac{π}{2} = 0 + 1 = 1$ Hence, the absolute maximum value of f is $\frac{5}{4}$ occurring at $x = \frac{π}{6}$ and the absolute minimum value of f is 1 occurring at $x = 0, \frac{π}{2}$ and $π$ .

Show that altitude of the right circular cone of maximum volume that can be inscribed in a sphere or radius r is $\frac{4 r}{3}$ . Sol. A sphere of fixed radius ( r ) is given. Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by, $V = \frac{1}{3} π R^{2} h$ In $△ BC D, BC = r^{2} - R^{2}$ $∴ h = r + r^{2} - R^{2} \Rightarrow R^{2} = 2 hr - h^{2}$ $∴ V = \frac{1}{3} π h (2 hr - h^{2}) = \frac{1}{3} π (2 h^{2} r - h^{3})$

$∴ \frac{dV}{dh} = \frac{1}{3} π (4 hr - 3 h^{2})$ and $\frac{d ^{2} V}{dh ^{2}} = \frac{1}{3} π (4 r - 6 h)$ for maxima or minima

$\frac{dV}{dh} = 0 \Rightarrow 4 hr - 3 h^{2} = 0 \Rightarrow h = 0 or h = \frac{4 r}{3}$

rejecting $h = 0 ∴ h = \frac{4 r}{3}$ when $h = \frac{4 r}{3}, \frac{d ^{2} V}{dh ^{2}} = \frac{1}{3} π (4 r - 8 r) = - \frac{4 r π}{3} < 0$ V is maximum when $h = \frac{4 r}{3}$ By second order derivative test, the volume of the cone is maximum when altitude of cone is $\frac{4 r}{3}$

Let f be a function defined on [a, b] such that $f^{'} (x) > 0$ , for all $x \in (a, b)$ . then prove that f is an increasing function on (a, b). Sol. Since, $f^{'} (x) > 0$ on (a, b) Then, f is a differentiable function on $(a, b)$ . Also, every differentiable function is continuous, Therefore, f is continuous on $[a, b]$ Let $x_{1}, x_{2} \in (a, b)$ such that, $x_{2} > x_{1}$ then by LMV theorem, there exists $c \in (a, b)$ s.t.

$f^{'} (c) = \frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}$

$\Rightarrow f (x_{2}) - f (x_{1}) = (x_{2} - x_{1}) f^{'} (c)$ $\Rightarrow f (x_{2}) - f (x_{1}) > 0$ as $x_{2} > x_{1}$ and $f^{'} (x) > 0$ $\Rightarrow f (x_{2}) > f (x_{1})$ $∴$ for $x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2})$ .

Therefore, f is an increasing function.

Show that the height of the cylinder of maximum volume that can be inscribed in sphere of radius R is $\frac{2 R}{3}$ . Also find the maximum volume. Sol. A sphere of fixed radius (R) is given. Let r and h be the radius and the height of the cylinder respectively. From the given figure, we have $h = 2 R^{2} - r^{2}$

The volume (V) of the cylinder is given by,

$V = π r^{2} h = 2 π r^{2} R^{2} - r^{2}$

$\Rightarrow V^{2} = 4 π^{2} r^{4} (R^{2} - r^{2}) = 4 π^{2} (R^{2} r^{4} - r^{6}) = Z$ (let) $∴ \frac{dZ}{dr} = 4 π^{2} (4 R^{2} r^{3} - 6 r^{5})$ and $\frac{d ^{2} Z}{dr ^{2}} = 4 π^{2} (12 R^{2} r^{2} - 30 r^{4})$ for maxima or minima $\frac{dZ}{dr} = 0$ $\Rightarrow 4 π^{2} (4 R^{2} r^{3} - 6 r^{5}) = 0$ $\Rightarrow r = 0$ or $r^{2} = \frac{2 R ^{2}}{3}$

rejecting $r = 0 ∴ r^{2} = \frac{2 R ^{2}}{3}$ at $r^{2} = \frac{2 R ^{2}}{3}$

$\frac{d ^{2} Z}{dr ^{2}} = 4 π^{2} [12 R^{2} (\frac{2 R ^{2}}{3}) - 30 (\frac{4 R ^{4}}{9})] = - \frac{64 R ^{4}}{3} < 0$

The volume is the maximum when $r^{2} = \frac{2 R ^{2}}{3}$ . When $r^{2} = \frac{2 R ^{2}}{3}$ , the height of the cylinder is

$2 R^{2} - \frac{2 R ^{2}}{3} = 2 \frac{R ^{2}}{3} = \frac{2 R}{3} .$

Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2 R}{3}$ . Also maximum volume

$= π r^{2} h = π \times \frac{2 R ^{2}}{3} \times \frac{2 R}{3} = \frac{4}{3 3} π R^{3}$

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle ' $α$ ' is one-third of the cone and the greatest volume of cylinder is $\frac{4}{27} π h^{3} tan^{2} α$ . Sol. The given right circular of fixed height (h) and semi-vertical angle $(α)$ can be drawn as :

Let the radius and height of cylinder be R and H respectively and radius of cone be r. Then, $∠ GAO = α, OG = r, OA = h, OE = R$ , and $CE = H$ . We have, In $Δ AO^{'} C$

$tan α = \frac{R}{h - H}$

$\Rightarrow R = tan α (h - H)$

Now, the volume (V) of the cylinder is given by,

$V = π R^{2} H V = π [tan α (h - H)]^{2} H [From equation (1)] V = π tan^{2} α \cdot [(h - H)^{2} \cdot H] \frac{dV}{dH} = π tan^{2} α [(h - H)^{2} \cdot 1 + H \cdot 2 (h - H) \times (- 1)] \frac{dV}{dH} = π tan^{2} α (h - H) (h - 3 H)$

For maxima or minima $\frac{dV}{dH} = 0$

$∴ π tan^{2} α (h - H) (h - 3 H) = 0 h - 3 H = 0 \Rightarrow H = \frac{h}{3} [∵ π tan^{2} α (h - H) \neq = 0]$

Also,

$\frac{d ^{2} V}{dH ^{2}} \Rightarrow \frac{d ^{2} V}{dH ^{2}} At H = π tan^{2} α \cdot [(h - H) \cdot (- 3) + (h - 3 H) \cdot (- 1)] = π tan^{2} α \cdot [- 4 h + 6 H] = \frac{h}{3}; \frac{d ^{2} V}{dH ^{2}} = π tan^{2} α \cdot (- 2 h) < 0$

By second order derivative test, the volume of the cylinder is the greatest when $H = \frac{h}{3}$ Now, $R = tan α (h - \frac{h}{3}) = \frac{2 h}{3} tan α$ Thus, the height of the cylinder is one-third the height of cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder can be obtained as:

$V = π R^{2} H = π (\frac{2 h}{3} tan α)^{2} (\frac{h}{3}) = π (\frac{4 h ^{2}}{9} tan^{2} α) (\frac{h}{3}) = \frac{4}{27} π h^{3} tan^{2} α$

Hence, the given result is proved. Choose the correct answer in the following question.

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of the wheat is increasing at the rate of ? (A) 1 m/h (B) $0.1 m / h$ (C) 1.1 m/h (D) 0.5 m/h

Sol. (A) Let r be the radius of the cylinder Then, volume (V) of the cylinder is given by $= π r^{2} h$ ,

$V = π (radius)^{2} \times height = π (10)^{2} h = 100 π h (radius = 10 m)$

Differentiating with respect to time t, we have

$\Rightarrow \frac{dV}{dt} = 100 π \frac{dh}{dt}$

The tank is being filled with wheat at the rate of 314 cubic metres per hour.

$\frac{dV}{dt} = 314 m^{3} / h$

Thus, we have :

$314 = 100 π \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{314}{100 ( 3.14 )} = \frac{314}{314} = 1$

Hence, the depth of wheat is increasing at the rate of 1 m/h. The correct answer is A.

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Check NCERT Solutions for Class 12 Maths for all chapters, with exercise-wise answers, useful formulas, and clear solutions that make Maths practice simpler and more effective.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 6 Application of Derivatives Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 6.1 Solutions	18 Questions & Solutions	Rate of change, instantaneous rate of change and average rate of change
Exercise 6.2 Solutions	19 Questions & Solutions	Increasing and decreasing functions and intervals of increase or decrease
Exercise 6.3 Solutions	29 Questions & Solutions	Maximum and minimum values of functions and optimisation problems
Miscellaneous Exercise Solutions	16 Questions & Solutions	Mixed questions based on applications of derivatives

5.0Key Features and Benefits of NCERT Solutions Class 12 Maths Chapter 6 Applications of Derivatives

Clear explanations help students understand how derivatives are applied to different mathematical and practical problems.
Step-by-step solutions make questions based on tangents, normals, increasing and decreasing functions, and maxima and minima easier to solve.
The solutions cover the exercises and questions included in the NCERT Class 12 Maths textbook.
Regular practice can help students improve their accuracy, problem-solving skills, and confidence while solving calculus questions.
A strong understanding of applications of derivatives can support preparation for mathematics olympiads and other competitive entrance examinations.
Clear concepts from this chapter provide a useful foundation for studying further topics in calculus and their applications.

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Frequently Asked Questions

Where can I find NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives?

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What should I know about tangents and normals from NCERT Class 12 Maths Chapter 6?

How are maxima and minima found using NCERT Class 12 Maths Chapter 6?

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NCERT Solutions Class 12 Maths Chapter 6 Application of Derivatives

1.0Key Concepts of NCERT Solutions Class 12 Maths Chapter 6 Applications of Derivatives

2.0NCERT Class 12 Maths Chapter 6 Applications of Derivatives : Detailed Solutions

EXERCISE - 6.1

EXERCISE - 6.2

EXERCISE - 6.3

MISCELLANEOUS EXERCISE

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 6 Application of Derivatives Exercise-wise Solutions

5.0Key Features and Benefits of NCERT Solutions Class 12 Maths Chapter 6 Applications of Derivatives

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